Integration by substitution In a nutshell Integration by substitution is a more concrete version of the reverse chain rule that makes it easier to integrate a wide variety of functions.
Indefinite integration by substitution Integration by substitution relies on introducing a new variable, u ( x ) u(x) u ( x )
procedure 1.
If not given one, pick a suitable function to write as u ( x ) u(x) u ( x )
2.
Differentiate u u u x x x d x dx d x
3.
Rewrite the integral so that only u u u d u du d u
4.
Evaluate the integral, which will result in a function of u u u
5.
Rewrite the answer to give a function of x x x
Example 1 Use the substitution u = x 2 + 1 u=x^2+1 u = x 2 + 1 to evaluate ∫ 7 x ( x 2 + 1 ) 2 d x \int \dfrac{7x}{(x^2+1)^2}\, dx ∫ ( x 2 + 1 ) 2 7 x d x .
Differentiate u u u d u d x \dfrac{du}{dx} d x d u d x dx d x
u = x 2 + 1 d u d x = 2 x d u = 2 x d x d x = 1 2 x d u \begin{aligned} u&=x^2+1\\\dfrac{du}{dx}&=2x\\du&=2x \, dx\\dx&=\dfrac{1}{2x}\,du \end{aligned} u d x d u d u d x = x 2 + 1 = 2 x = 2 x d x = 2 x 1 d u
Substitute this into the integral, simplifying where applicable:
∫ 7 x ( x 2 + 1 ) 2 d x = ∫ 7 x ( x 2 + 1 ) 2 ( 1 2 x d u ) = ∫ 7 2 ( x 2 + 1 ) 2 d u \begin{aligned} \int \dfrac{7x}{(x^2+1)^2}\,dx&=\int \dfrac{7\cancel{x}}{(x^2+1)^2}\, \left(\dfrac{1}{2 \cancel{x}}\,du \right)\\&=\int \dfrac{7}{2(x^2+1)^2}\,du \end{aligned} ∫ ( x 2 + 1 ) 2 7 x d x = ∫ ( x 2 + 1 ) 2 7 x ( 2 x 1 d u ) = ∫ 2 ( x 2 + 1 ) 2 7 d u
Use substitution to ensure the integral is only in terms of u u u
∫ 7 2 ( x 2 + 1 ) 2 d u = ∫ 7 2 ( u ) 2 d u = ∫ 7 2 u 2 d u \begin{aligned}\int \dfrac{7}{2(x^2+1)^2}\,du &=\int \dfrac{7}{2(u)^2}\,du\\ &=\int \dfrac{7}{2u^2}\,du\end{aligned} ∫ 2 ( x 2 + 1 ) 2 7 d u = ∫ 2 ( u ) 2 7 d u = ∫ 2 u 2 7 d u
Perform the integration with u u u
∫ 7 2 u 2 d u = ∫ 7 2 u − 2 d u = − 7 2 u − 1 + C = − 7 2 u + C \begin{aligned} \int \dfrac{7}{2u^2}\,du &=\int \dfrac72 u^{-2}\,du \\&=-\dfrac72u^{-1}+C\\&=-\dfrac{7}{2u}+C \end{aligned} ∫ 2 u 2 7 d u = ∫ 2 7 u − 2 d u = − 2 7 u − 1 + C = − 2 u 7 + C
Use substitution to rewrite the answer in terms of x x x
u = x 2 + 1 − 7 2 u + C = − 7 2 ( x 2 + 1 ) + C \begin{aligned} u&=x^2+1\\ -\dfrac{7}{2u}+C&=-\dfrac{7}{2(x^2+1)}+C \end{aligned} u − 2 u 7 + C = x 2 + 1 = − 2 ( x 2 + 1 ) 7 + C
∫ 7 x ( x 2 + 1 ) 2 d x = − 7 2 ( x 2 + 1 ) + C ‾ \int \dfrac{7x}{(x^2+1)^2}\,dx=\underline{-\dfrac{7}{2(x^2+1)}+C} ∫ ( x 2 + 1 ) 2 7 x d x = − 2 ( x 2 + 1 ) 7 + C
Definite integration by substitution Using integration by substitution on a definite integral has a very similar process to that of an indefinite integral. The key difference is that the limits also change with the variable.
Example 2 Use the substitution u = e x + 1 u=e^x+1 u = e x + 1 ∫ 0 ln ( 3 ) e 2 x e x + 1 d x \int_0^{\ln(3)} e^{2x}\sqrt{e^x+1}\,dx ∫ 0 l n ( 3 ) e 2 x e x + 1 d x
First, differentiate u u u d x dx d x
u = e x + 1 d u d x = e x d u = e x d x d x = 1 e x d u \begin{aligned} u&=e^x+1\\\dfrac{du}{dx}&=e^x \\du&=e^x \, dx\\dx&=\dfrac{1}{e^x}\,du \end{aligned} u d x d u d u d x = e x + 1 = e x = e x d x = e x 1 d u
Before substituting, find the corresponding limits for u u u
u = e x + 1 x = 0 → u = e 0 + 1 → u = 2 x = ln ( 3 ) → u = e ln ( 3 ) + 1 → u = 4 \space \space u=e^x+1\\ \begin{aligned}x=0&\rightarrow u=e^0+1\rightarrow u=2\\x=\ln(3) &\rightarrow u =e^{\ln(3)}+1\rightarrow u=4 \end{aligned} u = e x + 1 x = 0 x = ln ( 3 ) → u = e 0 + 1 → u = 2 → u = e l n ( 3 ) + 1 → u = 4
Rewrite the integral in terms of u u u
∫ 0 ln ( 3 ) e 2 x e x + 1 d x = ∫ 2 4 e 2 x e x + 1 ( 1 e x d u ) = ∫ 2 4 e x e x + 1 d u = ∫ 2 4 ( u − 1 ) u d u = ∫ 2 4 ( u u − u ) d u \begin{aligned} \int_0^{\ln(3)} e^{2x}\sqrt{e^x+1}\,dx&=\int_2^4e^{2x}\sqrt{e^x+1}\,\left(\dfrac{1}{e^x}\,du\right)\\&=\int_2^4 e^x\sqrt{e^x+1}\,du\\ &=\int_2^4(u-1)\sqrt{u}\,du\\&=\int_2^4 (u\sqrt u - \sqrt u)\, du\end{aligned} ∫ 0 l n ( 3 ) e 2 x e x + 1 d x = ∫ 2 4 e 2 x e x + 1 ( e x 1 d u ) = ∫ 2 4 e x e x + 1 d u = ∫ 2 4 ( u − 1 ) u d u = ∫ 2 4 ( u u − u ) d u
Compute the transformed integral:
∫ 2 4 ( u u − u ) d u = ∫ 2 4 ( u 3 2 − u 1 2 ) d u = [ 2 5 u 5 2 − 2 3 u 3 2 ] 2 4 = ( 2 5 ( 4 ) 5 2 − 2 3 ( 4 ) 3 2 ) − ( 2 5 ( 2 ) 5 2 − 2 3 ( 2 ) 3 2 ) = ( 2 5 ( 32 ) − 2 3 ( 8 ) ) − ( 2 5 ( 4 2 ) − 2 3 ( 2 2 ) ) = 112 15 − 4 15 2 \begin{aligned}\int_2^4 (u\sqrt u - \sqrt u)\, du &= \int _2^4 \left(u^\frac32 - u^\frac12\right)\,du\\&=\left[\dfrac25 u^\frac52 - \dfrac23 u^\frac32\right]_2^4\\&=\left(\dfrac25 (4)^\frac52 - \dfrac23 (4)^\frac32\right)-\left(\dfrac25 (2)^\frac52 - \dfrac23 (2)^\frac32\right)\\&=\left(\dfrac25 (32) - \dfrac23 (8)\right) - \left(\dfrac25 (4\sqrt{2}) - \dfrac23 (2\sqrt2)\right)\\&=\dfrac{112}{15}-\dfrac{4}{15}\sqrt2 \end{aligned} ∫ 2 4 ( u u − u ) d u = ∫ 2 4 ( u 2 3 − u 2 1 ) d u = [ 5 2 u 2 5 − 3 2 u 2 3 ] 2 4 = ( 5 2 ( 4 ) 2 5 − 3 2 ( 4 ) 2 3 ) − ( 5 2 ( 2 ) 2 5 − 3 2 ( 2 ) 2 3 ) = ( 5 2 ( 32 ) − 3 2 ( 8 ) ) − ( 5 2 ( 4 2 ) − 3 2 ( 2 2 ) ) = 15 112 − 15 4 2
∫ 0 ln ( 3 ) e 2 x e x + 1 d x = 4 15 ( 28 − 2 ) ‾ \int_0^{\ln(3)} e^{2x}\sqrt{e^x+1}\,dx=\underline{\dfrac{4}{15} \left(28-\sqrt2\right)} ∫ 0 l n ( 3 ) e 2 x e x + 1 d x = 15 4 ( 28 − 2 )
