Integration - practice questions

Integration - practice questions

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 1 / 22

The presentation is structured as follows. You’re given an integral. You should try and solve it.

If you struggle, then there’ll be a hint - usually an indication of the method you should use. Finally a full solution will be given.

The presentation is structured as follows. You’re given an integral. You should try and solve it. If you struggle, then there’ll be a hint - usually an indication of the method you should use.

Finally a full solution will be given.

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 2 / 22

The presentation is structured as follows. You’re given an integral. You should try and solve it. If you struggle, then there’ll be a hint - usually an indication of the method you should use. Finally a full solution will be given.

Example 1 - easy

Find Z

xp

x²+ 1 dx

Hint: use substitution u = x²+ 1.

Solution: If u = x²+ 1, then du

dx = 2x , which gives dx = 1 2x du. The integral becomes:

Z xp

x²+ 1 dx = Z

x√ u 1

2x du = 1 2

Z √ u du =

= 1

3u³² + c = 1 3u√

u + c = 1

3(x²+ 1)p

x²+ 1 + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 3 / 22

Example 1 - easy

Find Z

xp

x²+ 1 dx

Hint:

use substitution u = x²+ 1.

Solution: If u = x²+ 1, then du

dx = 2x , which gives dx = 1 2x du. The integral becomes:

Z xp

x²+ 1 dx = Z

x√ u 1

2x du = 1 2

Z √ u du =

= 1

3u³² + c = 1 3u√

u + c = 1

3(x²+ 1)p

x²+ 1 + c

Example 1 - easy

Find Z

xp

x²+ 1 dx

Hint: use substitution u = x²+ 1.

Solution: If u = x²+ 1, then du

dx = 2x , which gives dx = 1 2x du. The integral becomes:

Z xp

x²+ 1 dx = Z

x√ u 1

2x du = 1 2

Z √ u du =

= 1

3u³² + c = 1 3u√

u + c = 1

3(x²+ 1)p

x²+ 1 + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 3 / 22

Example 1 - easy

Find Z

xp

x²+ 1 dx

Hint: use substitution u = x²+ 1.

Solution:

If u = x²+ 1, then du

dx = 2x , which gives dx = 1 2x du. The integral becomes:

Z xp

x²+ 1 dx = Z

x√ u 1

2x du = 1 2

Z √ u du =

= 1

3u³² + c = 1 3u√

u + c = 1

3(x²+ 1)p

x²+ 1 + c

Example 1 - easy

Find Z

xp

x²+ 1 dx

Hint: use substitution u = x²+ 1.

Solution: If u = x²+ 1, then du

dx = 2x , which gives dx = 1 2x du.

The integral becomes:

Z xp

x²+ 1 dx = Z

x√ u 1

2x du = 1 2

Z √ u du =

= 1

3u³² + c = 1 3u√

u + c = 1

3(x²+ 1)p

x²+ 1 + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 3 / 22

Example 2 - easy

Find Z

x√

3x − 1 dx

Hint: use substitution u = 3x − 1. Solution: If u = 3x − 1, then du

dx = 3, which gives dx = 1 3 du. We also have x = u + 1

3 . The integral becomes:

Z x√

3x − 1 dx =

Z u + 1 3

√u1

3 du = 1 9

Z u√

u +√ u du =

= 2

45u⁵² + 2

27u³² + c = 2

45(3x − 1)⁵² + 2

27(3x − 1)³² + c

Example 2 - easy

Find Z

x√

3x − 1 dx Hint:

use substitution u = 3x − 1. Solution: If u = 3x − 1, then du

dx = 3, which gives dx = 1 3 du. We also have x = u + 1

3 . The integral becomes:

Z x√

3x − 1 dx =

Z u + 1 3

√u1

3 du = 1 9

Z u√

u +√ u du =

= 2

45u⁵² + 2

27u³² + c = 2

45(3x − 1)⁵² + 2

27(3x − 1)³² + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 4 / 22

Example 2 - easy

Find Z

x√

3x − 1 dx

Hint: use substitution u = 3x − 1.

Solution: If u = 3x − 1, then du

dx = 3, which gives dx = 1 3 du. We also have x = u + 1

3 . The integral becomes:

Z x√

3x − 1 dx =

Z u + 1 3

√u1

3 du = 1 9

Z u√

u +√ u du =

= 2

45u⁵² + 2

27u³² + c = 2

45(3x − 1)⁵² + 2

27(3x − 1)³² + c

Example 2 - easy

Find Z

x√

3x − 1 dx

Hint: use substitution u = 3x − 1.

Solution:

If u = 3x − 1, then du

dx = 3, which gives dx = 1 3 du. We also have x = u + 1

3 . The integral becomes:

Z x√

3x − 1 dx =

Z u + 1 3

√u1

3 du = 1 9

Z u√

u +√ u du =

= 2

45u⁵² + 2

27u³² + c = 2

45(3x − 1)⁵² + 2

27(3x − 1)³² + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 4 / 22

Example 2 - easy

Find Z

x√

3x − 1 dx

Hint: use substitution u = 3x − 1.

Solution: If u = 3x − 1, then du

dx = 3, which gives dx = 1 3 du.

We also have x = u + 1 3 . The integral becomes:

Z x√

3x − 1 dx =

Z u + 1 3

√u1

3 du = 1 9

Z u√

u +√ u du =

= 2

45u⁵² + 2

27u³² + c = 2

45(3x − 1)⁵² + 2

27(3x − 1)³² + c

Example 3 - easy

Find Z

x⁴ln x dx

Hint: use integration by parts with f = ln x and g⁰ = x⁴. Solution: If f = ln x , then f⁰ = 1

x. Also if g⁰ = x⁴, then g = ¹₅x⁵. The integral becomes:

Z

x⁴ln x dx = 1

5x⁵ln x − Z 1

x ×1

5x⁵ dx = 1

5x⁵ln x −1 5

Z

x⁴dx =

= 1

5x⁵ln x − 1 25x⁵+ c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 5 / 22

Example 3 - easy

Find Z

x⁴ln x dx

Hint:

use integration by parts with f = ln x and g⁰ = x⁴. Solution: If f = ln x , then f⁰ = 1

x. Also if g⁰ = x⁴, then g = ¹₅x⁵. The integral becomes:

Z

x⁴ln x dx = 1

5x⁵ln x − Z 1

x ×1

5x⁵ dx = 1

5x⁵ln x −1 5

Z

x⁴dx =

= 1

5x⁵ln x − 1 25x⁵+ c

Example 3 - easy

Find Z

x⁴ln x dx

Hint: use integration by parts with f = ln x and g⁰= x⁴.

Solution: If f = ln x , then f⁰ = 1

x. Also if g⁰ = x⁴, then g = ¹₅x⁵. The integral becomes:

Z

x⁴ln x dx = 1

5x⁵ln x − Z 1

x ×1

5x⁵ dx = 1

5x⁵ln x −1 5

Z

x⁴dx =

= 1

5x⁵ln x − 1 25x⁵+ c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 5 / 22

Example 3 - easy

Find Z

x⁴ln x dx

Hint: use integration by parts with f = ln x and g⁰= x⁴. Solution:

If f = ln x , then f⁰ = 1

x. Also if g⁰ = x⁴, then g = ¹₅x⁵. The integral becomes:

Z

x⁴ln x dx = 1

5x⁵ln x − Z 1

x ×1

5x⁵ dx = 1

5x⁵ln x −1 5

Z

x⁴dx =

= 1

5x⁵ln x − 1 25x⁵+ c

Example 3 - easy

Find Z

x⁴ln x dx

Hint: use integration by parts with f = ln x and g⁰= x⁴. Solution: If f = ln x , then f⁰ = 1

x. Also if g⁰ = x⁴, then g = ¹₅x⁵. The integral becomes:

Z

x⁴ln x dx = 1

5x⁵ln x − Z 1

x ×1

5x⁵ dx = 1

5x⁵ln x −1 5

Z

x⁴dx =

= 1

5x⁵ln x − 1 25x⁵+ c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 5 / 22

Example 4 - easy

Find

Z x

x²− 1 dx

Hint: the denominator can be factorized, so you can try partial fractions, but it’s much better to look for the derivative of the denominator in the numerator.

Solution: the derivative of the denominator is 2x , so this is what we want in the numerator:

Z x

x²− 1 dx = 1 2

Z 2x

x²− 1 dx = 1

2ln |x²− 1| + c

Example 4 - easy

Find

Z x

x²− 1 dx Hint:

the denominator can be factorized, so you can try partial fractions, but it’s much better to look for the derivative of the denominator in the numerator.

Solution: the derivative of the denominator is 2x , so this is what we want in the numerator:

Z x

x²− 1 dx = 1 2

Z 2x

x²− 1 dx = 1

2ln |x²− 1| + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 6 / 22

Example 4 - easy

Find

Z x

x²− 1 dx

Hint: the denominator can be factorized, so you can try partial fractions, but it’s much better to look for the derivative of the denominator in the numerator.

Solution: the derivative of the denominator is 2x , so this is what we want in the numerator:

Z x

x²− 1 dx = 1 2

Z 2x

x²− 1 dx = 1

2ln |x²− 1| + c

Example 4 - easy

Find

Z x

x²− 1 dx

Hint: the denominator can be factorized, so you can try partial fractions, but it’s much better to look for the derivative of the denominator in the numerator.

Solution:

the derivative of the denominator is 2x , so this is what we want in the numerator:

Z x

x²− 1 dx = 1 2

Z 2x

x²− 1 dx = 1

2ln |x²− 1| + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 6 / 22

Example 4 - easy

Find

Z x

x²− 1 dx

Hint: the denominator can be factorized, so you can try partial fractions, but it’s much better to look for the derivative of the denominator in the numerator.

Solution: the derivative of the denominator is 2x , so this is what we want in the numerator:

Z x

x²− 1 dx = 1 2

Z 2x

x²− 1 dx = 1

2ln |x²− 1| + c

Example 5 - easy

Find

Z 4x − 5 6x²− x − 2 dx

Hint: we can factorize the denominator, so partial fractions can be used. Solution: 6x²− x − 2 = (3x − 2)(2x + 1), so we write:

4x − 5

6x²− x − 2 = A

3x − 2 + B

2x + 1 = A(2x + 1) + B(3x − 2) (3x − 2)(2x + 1)

Equating the coefficients of x in the numerator gives the following system of equations:

(2A + 3B = 4 A − 2B = −5 Solving the above gives A = −1 and B = 2.

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 7 / 22

Example 5 - easy

Find

Z 4x − 5 6x²− x − 2 dx Hint:

we can factorize the denominator, so partial fractions can be used. Solution: 6x²− x − 2 = (3x − 2)(2x + 1), so we write:

4x − 5

6x²− x − 2 = A

3x − 2 + B

2x + 1 = A(2x + 1) + B(3x − 2) (3x − 2)(2x + 1)

Equating the coefficients of x in the numerator gives the following system of equations:

(2A + 3B = 4 A − 2B = −5 Solving the above gives A = −1 and B = 2.

Example 5 - easy

Find

Z 4x − 5 6x²− x − 2 dx

Hint: we can factorize the denominator, so partial fractions can be used.

Solution: 6x²− x − 2 = (3x − 2)(2x + 1), so we write: 4x − 5

6x²− x − 2 = A

3x − 2 + B

2x + 1 = A(2x + 1) + B(3x − 2) (3x − 2)(2x + 1)

Equating the coefficients of x in the numerator gives the following system of equations:

(2A + 3B = 4 A − 2B = −5 Solving the above gives A = −1 and B = 2.

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 7 / 22

Example 5 - easy

Find

Z 4x − 5 6x²− x − 2 dx

Hint: we can factorize the denominator, so partial fractions can be used.

Solution:

6x²− x − 2 = (3x − 2)(2x + 1), so we write: 4x − 5

6x²− x − 2 = A

3x − 2 + B

2x + 1 = A(2x + 1) + B(3x − 2) (3x − 2)(2x + 1)

Equating the coefficients of x in the numerator gives the following system of equations:

(2A + 3B = 4 A − 2B = −5 Solving the above gives A = −1 and B = 2.

Example 5 - easy

Find

Z 4x − 5 6x²− x − 2 dx

Hint: we can factorize the denominator, so partial fractions can be used.

Solution: 6x²− x − 2 = (3x − 2)(2x + 1), so we write:

4x − 5

6x²− x − 2 = A

3x − 2 + B

2x + 1 = A(2x + 1) + B(3x − 2) (3x − 2)(2x + 1)

Equating the coefficients of x in the numerator gives the following system of equations:

(2A + 3B = 4 A − 2B = −5 Solving the above gives A = −1 and B = 2.

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 7 / 22

Example 5 continued

Z 4x − 5

6x²− x − 2 dx =

Z −1

3x − 2 + 2

2x + 1 dx =

= −

Z 1

3x − 2 dx + 2

Z 1

2x + 1 dx = −1

3ln |3x − 2| + ln |2x + 1| + c

Example 6 - easy

Find

Z 3

x²+ 2x + 6 dx

Hint: the denominator cannot be factorized, so we complete the square and look for arctan x formula.

Solution:

Z 3

x²+ 2x + 6 dx =

Z 3

(x + 1)²+ 5 dx = 3

√ 5

Z √

5

(x + 1)²+ 5 dx =

= 3

√

5arctan x + 1

√ 5

 + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 9 / 22

Example 6 - easy

Find

Z 3

x²+ 2x + 6 dx Hint:

the denominator cannot be factorized, so we complete the square and look for arctan x formula.

Solution:

Z 3

x²+ 2x + 6 dx =

Z 3

(x + 1)²+ 5 dx = 3

√ 5

Z √

5

(x + 1)²+ 5 dx =

= 3

√

5arctan x + 1

√ 5

 + c

Example 6 - easy

Find

Z 3

x²+ 2x + 6 dx

Hint: the denominator cannot be factorized, so we complete the square and look for arctan x formula.

Solution:

Z 3

x²+ 2x + 6 dx =

Z 3

(x + 1)²+ 5 dx = 3

√ 5

Z √

5

(x + 1)²+ 5 dx =

= 3

√

5arctan x + 1

√ 5

 + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 9 / 22

Example 6 - easy

Find

Z 3

x²+ 2x + 6 dx

Hint: the denominator cannot be factorized, so we complete the square and look for arctan x formula.

Solution:

Z 3

x²+ 2x + 6 dx =

Z 3

(x + 1)²+ 5 dx = 3

√ 5

Z √

5

(x + 1)²+ 5 dx =

= 3

√

5arctan x + 1

√ 5

 + c

Example 6 - easy

Find

Z 3

x²+ 2x + 6 dx

Hint: the denominator cannot be factorized, so we complete the square and look for arctan x formula.

Solution:

Z 3

x²+ 2x + 6 dx =

Z 3

(x + 1)²+ 5 dx = 3

√ 5

Z √

5

(x + 1)²+ 5 dx =

= 3

√

5arctan x + 1

√ 5

 + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 9 / 22

Example 7 - easy

Find Z

x²e^3x dx

Hint: you need to integrate by parts twice.

Solution: we let f = x² and g⁰ = e^3x first. Then we let f₂ = x and g₂⁰ = e^3x.

Z

x²e^3x dx = 1

3x²e^3x−2 3

Z

xe^3x dx = 1

3x²e^3x −2

9xe^3x +2 9

Z

e^3x dx =

= 1

3x²e^3x−2

9xe^3x + 2

27e^3x+ c

Example 7 - easy

Find Z

x²e^3x dx Hint:

you need to integrate by parts twice.

Solution: we let f = x² and g⁰ = e^3x first. Then we let f₂ = x and g₂⁰ = e^3x.

Z

x²e^3x dx = 1

3x²e^3x−2 3

Z

xe^3x dx = 1

3x²e^3x −2

9xe^3x +2 9

Z

e^3x dx =

= 1

3x²e^3x−2

9xe^3x + 2

27e^3x+ c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 10 / 22

Example 7 - easy

Find Z

x²e^3x dx

Hint: you need to integrate by parts twice.

Solution: we let f = x² and g⁰ = e^3x first. Then we let f₂ = x and g₂⁰ = e^3x.

Z

x²e^3x dx = 1

3x²e^3x−2 3

Z

xe^3x dx = 1

3x²e^3x −2

9xe^3x +2 9

Z

e^3x dx =

= 1

3x²e^3x−2

9xe^3x + 2

27e^3x+ c

Example 7 - easy

Find Z

x²e^3x dx

Hint: you need to integrate by parts twice.

Solution:

we let f = x² and g⁰ = e^3x first. Then we let f₂ = x and g₂⁰ = e^3x.

Z

x²e^3x dx = 1

3x²e^3x−2 3

Z

xe^3x dx = 1

3x²e^3x −2

9xe^3x +2 9

Z

e^3x dx =

= 1

3x²e^3x−2

9xe^3x + 2

27e^3x+ c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 10 / 22

Example 7 - easy

Find Z

x²e^3x dx

Hint: you need to integrate by parts twice.

Solution: we let f = x² and g⁰ = e^3x first. Then we let f2 = x and g₂⁰ = e^3x.

Z

x²e^3x dx = 1

3x²e^3x−2 3

Z

xe^3x dx = 1

3x²e^3x −2

9xe^3x +2 9

Z

e^3x dx =

= 1

3x²e^3x−2

9xe^3x + 2

27e^3x+ c

Example 8 - easy

Find Z

e^2xsin(x ) dx

Hint: you need to integrate by parts twice and use the resulting equation to find the integral.

Solution: we let f = e^2x and g⁰= sin x first. Then we let f₂= e^2x and g₂⁰ = cos x .

Z

e^2xsin(x ) dx = −e^2xcos x + 2 Z

e^2xcos(x ) dx =

= −e^2xcos x + 2e^2xsin x − 4 Z

e^2xsin(x ) dx

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 11 / 22

Example 8 - easy

Find Z

e^2xsin(x ) dx Hint:

you need to integrate by parts twice and use the resulting equation to find the integral.

Solution: we let f = e^2x and g⁰= sin x first. Then we let f₂= e^2x and g₂⁰ = cos x .

Z

e^2xsin(x ) dx = −e^2xcos x + 2 Z

e^2xcos(x ) dx =

= −e^2xcos x + 2e^2xsin x − 4 Z

e^2xsin(x ) dx

Example 8 - easy

Find Z

e^2xsin(x ) dx

Hint: you need to integrate by parts twice and use the resulting equation to find the integral.

Solution: we let f = e^2x and g⁰= sin x first. Then we let f₂= e^2x and g₂⁰ = cos x .

Z

e^2xsin(x ) dx = −e^2xcos x + 2 Z

e^2xcos(x ) dx =

= −e^2xcos x + 2e^2xsin x − 4 Z

e^2xsin(x ) dx

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 11 / 22

Example 8 - easy

Find Z

e^2xsin(x ) dx

Hint: you need to integrate by parts twice and use the resulting equation to find the integral.

Solution:

we let f = e^2x and g⁰= sin x first. Then we let f₂= e^2x and g₂⁰ = cos x .

Z

e^2xsin(x ) dx = −e^2xcos x + 2 Z

e^2xcos(x ) dx =

= −e^2xcos x + 2e^2xsin x − 4 Z

e^2xsin(x ) dx

Example 8 - easy

Find Z

e^2xsin(x ) dx

Hint: you need to integrate by parts twice and use the resulting equation to find the integral.

Solution: we let f = e^2x and g⁰= sin x first. Then we let f₂= e^2x and g₂⁰ = cos x .

Z

e^2xsin(x ) dx = −e^2xcos x + 2 Z

e^2xcos(x ) dx =

= −e^2xcos x + 2e^2xsin x − 4 Z

e^2xsin(x ) dx

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 11 / 22

Example 8 continued

So we get that:

Z

e^2xsin(x ) dx = −e^2xcos x + 2e^2xsin x − 4 Z

e^2xsin(x ) dx Rearranging the equation we got and dividing by 5 gives:

Z

e^2xsin(x ) dx = −e^2xcos x + 2e^2xsin x

5 + c

Example 9 - moderate

Find

Z 1

e^x + 9e^−x dx

Hint: try substitution u = e^x and look for arctan x . Solution: if u = e^x, then du

dx = e^x, so dx = 1

e^xdu = 1 udu.

Z 1

e^x + 9e^−x dx =

Z 1

u + 9u⁻¹ ×1 u du =

Z 1

u²+ 9 du =

= 1 3

Z 3

u²+ 9 du = 1

3arctan u 3

 + c =

= 1

3arctan e^x 3

 + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 13 / 22

Example 9 - moderate

Find

Z 1

e^x + 9e^−x dx Hint:

try substitution u = e^x and look for arctan x . Solution: if u = e^x, then du

dx = e^x, so dx = 1

e^xdu = 1 udu.

Z 1

e^x + 9e^−x dx =

Z 1

u + 9u⁻¹ ×1 u du =

Z 1

u²+ 9 du =

= 1 3

Z 3

u²+ 9 du = 1

3arctan u 3

 + c =

= 1

3arctan e^x 3

 + c

Example 9 - moderate

Find

Z 1

e^x + 9e^−x dx

Hint: try substitution u = e^x and look for arctan x .

Solution: if u = e^x, then du

dx = e^x, so dx = 1

e^xdu = 1 udu.

Z 1

e^x + 9e^−x dx =

Z 1

u + 9u⁻¹ ×1 u du =

Z 1

u²+ 9 du =

= 1 3

Z 3

u²+ 9 du = 1

3arctan u 3

 + c =

= 1

3arctan e^x 3

 + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 13 / 22

Example 9 - moderate

Find

Z 1

e^x + 9e^−x dx

Hint: try substitution u = e^x and look for arctan x . Solution:

if u = e^x, then du

dx = e^x, so dx = 1

e^xdu = 1 udu.

Z 1

e^x + 9e^−x dx =

Z 1

u + 9u⁻¹ ×1 u du =

Z 1

u²+ 9 du =

= 1 3

Z 3

u²+ 9 du = 1

3arctan u 3

 + c =

= 1

3arctan e^x 3

 + c

Example 9 - moderate

Find

Z 1

e^x + 9e^−x dx

Hint: try substitution u = e^x and look for arctan x . Solution: if u = e^x, then du

dx = e^x, so dx = 1

e^xdu = 1 udu.

Z 1

e^x + 9e^−x dx =

Z 1

u + 9u⁻¹ ×1 u du =

Z 1

u²+ 9 du =

= 1 3

Z 3

u²+ 9 du = 1

3arctan u 3

 + c =

= 1

3arctan e^x 3

 + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 13 / 22

Example 10 - moderate

Find

Z x² 1 − x² dx

Hint: divide and then use partial fractions. Solution:

Z x²

1 − x² dx =

Z x²− 1 + 1 1 − x² dx =

Z

−1 + 1

1 − x² dx =

= Z

−1 +

1 2

1 − x +

1 2

1 + x dx =

= −x − 1

2ln |1 − x | +1

2ln |1 + x | + c

Example 10 - moderate

Find

Z x² 1 − x² dx Hint:

divide and then use partial fractions. Solution:

Z x²

1 − x² dx =

Z x²− 1 + 1 1 − x² dx =

Z

−1 + 1

1 − x² dx =

= Z

−1 +

1 2

1 − x +

1 2

1 + x dx =

= −x − 1

2ln |1 − x | +1

2ln |1 + x | + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 14 / 22

Example 10 - moderate

Find

Z x² 1 − x² dx

Hint: divide and then use partial fractions.

Solution:

Z x²

1 − x² dx =

Z x²− 1 + 1 1 − x² dx =

Z

−1 + 1

1 − x² dx =

= Z

−1 +

1 2

1 − x +

1 2

1 + x dx =

= −x − 1

2ln |1 − x | +1

2ln |1 + x | + c

Example 10 - moderate

Find

Z x² 1 − x² dx

Hint: divide and then use partial fractions.

Solution:

Z x²

1 − x² dx =

Z x²− 1 + 1 1 − x² dx =

Z

−1 + 1

1 − x² dx =

= Z

−1 +

1 2

1 − x +

1 2

1 + x dx =

= −x − 1

2ln |1 − x | +1

2ln |1 + x | + c

Tomasz Lechowski Batory 2IB A & A HL September 11, 2020 14 / 22

Example 10 - moderate

Find

Z x² 1 − x² dx

Hint: divide and then use partial fractions.

Solution:

Z x²

1 − x² dx =

Z x²− 1 + 1 1 − x² dx =

Z

−1 + 1

1 − x² dx =

= Z

−1 +

1 2

1 − x +

1 2

1 + x dx =

= −x − 1

2ln |1 − x | +1

2ln |1 + x | + c