Evaluate the indefinite integral: int x/(x^2+4)^3\,dx Use the substitution u = x^2 + 4.

Let u = x^2+4, so du = 2x\,dx, which gives x\,dx = 1/2du. Substituting into the integral, int x/(x^2+4)^3\,dx = int 1/u^3cdot1/2\,du = 1/2int u^-3\,du. Applying the power rule, 1/2int u^-3\,du = 1/2·fracu^-2-2 + C = -1/4u^-2+C. Substituting back u = x^2+4 gives the final answer -dfrac14(x^2+4)^2+C.

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Indefinite integral via u-substitution

Evaluate the indefinite integral:

$\int \frac{x}{(x^2+4)^3}\,dx$

Use the substitution $u = x^2 + 4$ .

Hint

Compute $du$ in terms of $x\,dx$ first. Notice that $x\,dx$ appears in the numerator, so you can replace it directly using $du = 2x\,dx$ . Then rewrite the integral entirely in terms of $u$ , integrate using the power rule, and substitute back for $x$ at the end.

Step-by-step solution

Let $u = x^2+4$ , so $du = 2x\,dx$ , which gives $x\,dx = \frac{1}{2}du$ . Substituting into the integral, $\int \frac{x}{(x^2+4)^3}\,dx = \int \frac{1}{u^3}\cdot\frac{1}{2}\,du = \frac{1}{2}\int u^{-3}\,du$ .

Applying the power rule, $\frac{1}{2}\int u^{-3}\,du = \frac{1}{2}\cdot\frac{u^{-2}}{-2} + C = -\frac{1}{4}u^{-2}+C$ . Substituting back $u = x^2+4$ gives the final answer $-\dfrac{1}{4(x^2+4)^2}+C$ .

Indefinite integral via u-substitution

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