How to Use the Formula for Integration by Parts

Integration by parts is simply the product rule reversed. The formula is as follows:

Formula

Integration by Parts

∫ uv′dx = uv −∫ u′vdx

Note! In exercises with integration by parts, you should choose ex as v′ and ln ⁡ (x) as u.

Example 1

∫ 3xexdx = ∗3xex −∫ 3exdx = 3xex − 3ex + C = 3ex(x − 1) + C

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u = 3xv′ = ex u′ = 3 v = ex

Example 2

Find the function F such that F′(x) = 4x3 + 1 x and F(1) = 2

F(x) = ∫ 4x3 + 1 xdx = x4 + ln ⁡ |x| + C

Furthermore, given that F(1) = 2:

2 = F(1) = 14 + ln ⁡ |1| + C = 1 + 0 + C, C = 1

Then, F(x) = x4 + ln ⁡ |x| + 1.

Example 3

Compute the integral ∫ ⁡ sin ⁡ 2xdx

∫ sin ⁡ 2xdx = ∫ sin ⁡ x ⋅ sin ⁡ xdx = ∗− sin ⁡ x cos ⁡ x = + ∫ cos ⁡ 2xdx = − sin ⁡ x cos ⁡ x = + ∫ 1 − sin ⁡ 2xdx = − sin ⁡ x cos ⁡ x = + x∫ sin ⁡ 2xdx

∫ sin ⁡ 2xdx = ∫ sin ⁡ x ⋅ sin ⁡ xdx = ∗− sin ⁡ x cos ⁡ x + ∫ cos ⁡ 2xdx = − sin ⁡ x cos ⁡ x + ∫ 1 − sin ⁡ 2xdx = − sin ⁡ x cos ⁡ x + x −∫ sin ⁡ 2xdx

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u = sin ⁡ xv′ = sin ⁡ x u′ = cos ⁡ xv = − cos ⁡ x

This gives you an equation that you solve for ∫ ⁡ sin ⁡ 2xdx:

∫ sin ⁡ 2xdx = − sin ⁡ x cos ⁡ x = + x∫ sin ⁡ 2xdx

2 ⋅∫ sin ⁡ 2xdx 2 = − sin ⁡ x cos ⁡ x + x 2 ∫ sin ⁡ 2xdx = −1 2 sin ⁡ x cos ⁡ x + x 2 + C

∫ sin ⁡ 2xdx = − sin ⁡ x cos ⁡ x + x −∫ sin ⁡ 2xdx 2 ⋅∫ sin ⁡ 2xdx = − sin ⁡ x cos ⁡ x + x| : 2 ∫ sin ⁡ 2xdx = −1 2 sin ⁡ x cos ⁡ x + x 2 + C

Example 4

Compute ∫ ⁡ cos ⁡ (2x) sin ⁡ (2x)dx

= ∫ cos ⁡ (2x) sin ⁡ (2x)dx = ∗− 1 2 cos ⁡ 2(2x) −∫ sin ⁡ (2x) cos ⁡ (2x)dx

∫ cos ⁡ (2x) sin ⁡ (2x)dx = ∗−1 2 cos ⁡ 2(2x) −∫ sin ⁡ (2x) cos ⁡ (2x)dx

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u = cos ⁡ (2x) v′ = sin ⁡ (2x) u′ = −2 sin ⁡ (2x)v = −1 2 cos ⁡ (2x)

You now solve this expression as an equation with respect to ∫ ⁡ cos ⁡ (2x) sin ⁡ (2x)dx:

∫ cos ⁡ (2x) sin ⁡ (2x)dx = −1 2 cos ⁡ 2(2x) − ∫ sin ⁡ (2x) ⋅cos ⁡ (2x)dx 2 ∫ cos ⁡ (2x) sin ⁡ (2x)dx = −1 2 cos ⁡ 2(2x)| ÷2 ∫ cos ⁡ (2x) sin ⁡ (2x)dx = −1 4 cos ⁡ 2(2x) + C

∫ cos ⁡ (2x) sin ⁡ (2x)dx = −1 2 cos ⁡ 2(2x) −∫ sin ⁡ (2x) cos ⁡ (2x)dx 2 ∫ cos ⁡ (2x) sin ⁡ (2x)dx = −1 2 cos ⁡ 2(2x)| ÷ 2 ∫ cos ⁡ (2x) sin ⁡ (2x)dx = −1 4 cos ⁡ 2(2x) + C

Example 5

Compute ∫ ⁡ex (x2 + 3x − 4) dx

= ∫ ex (x2 + 3x − 4) dx = ∗ex (x2 + 3x − 4) −∫ ex(2x + 3)dx = ex (x2 + 3x − 4) −(ex(2x + 3) −∫ 2exdx ) = ∗∗ex (x2 + 3x − 4) − ex(2x + 3) + 2ex + C = ex (x2 + x − 5) + C

∫ ex (x2 + 3x − 4) dx = ∗ex (x2 + 3x − 4) −∫ ex(2x + 3)dx = ex (x2 + 3x − 4) −(ex(2x + 3) −∫ 2exdx) = ∗∗ex (x2 + 3x − 4) − ex(2x + 3) + 2ex + C = ex (x2 + x − 5) + C

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u = x2 + 3x − 4v′ = ex u′ = 2x + 3 v = ex

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z = 2x + 3w′ = ex z′ = 2 w = ex