8.5 Integration by substitution and by parts — Learn

Prerequisites: the chain rule and the product rule from Section 7.7; standard integrals and the power rule for integration from Section 8.1.

By the end of this section you can: use one substitution to turn an integral into a standard form; change the limits of a definite integral when you substitute; integrate a product by parts; and choose between the two methods.

1. Substitution reverses the chain rule

The chain rule differentiates a function of a function. Substitution runs it backwards. If $u = g(x)$, then

and the integral becomes

Choose $u$ to be the inside function whose derivative also appears in the integral. Then replace every part of the integral, including $dx$, and finish with a standard integral.

Example 1. Find $\displaystyle\int 2x(x^2 + 1)^3dx$.

Let $u = x^2 + 1$. Then $du = 2xdx$, so $2xdx = du$. The integral becomes

Substitute back:

Example 2. Find $\displaystyle\int 3x^2e^{x^3}dx$.

Let $u = x^3$, so $du = 3x^2dx$. Then

Example 3. Find $\displaystyle\int \frac{2x}{x^2 + 1}dx$.

Let $u = x^2 + 1$, so $du = 2xdx$. Then

Definite integrals: change the limits

For a definite integral, change the limits to values of $u$ as you substitute. Then you never need to return to $x$.

Example 4. Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^3dx$.

Let $u = x^2 + 1$, so $du = 2xdx$. Change the limits:

• when $x = 0$, $u = 0^2 + 1 = 1$;
• when $x = 1$, $u = 1^2 + 1 = 2$.

Then

2. Integration by parts reverses the product rule

The product rule differentiates a product. Integration by parts runs it backwards. For two functions $u$ and $v$,

In the shorter form, with $dv = v'(x)dx$ and $du = u'(x)dx$,

Choose $u$ to be the part that becomes simpler when differentiated. A useful order to try is logarithm, inverse trigonometric, algebraic, trigonometric, exponential. The first type on that list is usually the best choice for $u$.

Example 5. Find $\displaystyle\int xe^{2x}dx$.

Take $u = x$ and $\dfrac{dv}{dx} = e^{2x}$. Then $\dfrac{du}{dx} = 1$ and $v = \dfrac{1}{2}e^{2x}$. So

Taking out the common factor $e^{2x}$ gives

Example 6. Find $\displaystyle\int x\sin xdx$.

Take $u = x$ and $\dfrac{dv}{dx} = \sin x$. Then $\dfrac{du}{dx} = 1$ and $v = -\cos x$. So

Example 7. Find $\displaystyle\int \ln xdx$.

Write the integrand as $\ln x \times 1$. Take $u = \ln x$ and $\dfrac{dv}{dx} = 1$. Then $\dfrac{du}{dx} = \dfrac{1}{x}$ and $v = x$. So

Example 8. Find $\displaystyle\int x^2e^xdx$.

Take $u = x^2$ and $\dfrac{dv}{dx} = e^x$. Then $\dfrac{du}{dx} = 2x$ and $v = e^x$, so

The new integral needs parts again, with $u = x$ and $v = e^x$:

Substitute this back:

Example 9. Evaluate $\displaystyle\int_0^1 xe^xdx$.

By parts with $u = x$ and $v = e^x$,

To evaluate the definite integral, put the limits on the whole antiderivative:

Watch out. Four mistakes are common.

Not replacing $dx$. When you substitute $u = g(x)$, you must replace $dx$ as well. If any $x$ or $dx$ remains, the substitution is unfinished. Write $du = g'(x)dx$ and use it.

Forgetting to change the limits. For a definite integral, convert the $x$-limits to $u$-limits before integrating. Leaving the old limits in place gives the wrong value.

Choosing the wrong part for $u$ in by parts. If the new integral is harder than the first, swap the choice. Putting the polynomial as $u$ makes it simpler; putting the exponential as $u$ does not.

Dropping the minus sign in by parts. The formula is $uv - \int vdu$, with a minus. Re-check: differentiate your answer and compare it with the original integrand.

You should now be able to

• choose a substitution and find $du$;
• rewrite an integral completely in terms of $u$;
• change the limits of a definite integral when substituting;
• apply the integration by parts formula;
• use integration by parts twice when a power of $x$ is multiplied by an exponential;
• check an integral by differentiating the answer.