Math · Calculus I · Worked example

Integrate 2x(x² + 1)⁵ by substitution

Find ∫2x(x² + 1)⁵ dx.

\int 2 x {(x^{2} + 1)}^{5} d x

Choose u

The inner function is x² + 1, and its derivative 2x is also a factor. Let u = x² + 1.

u = x^{2} + 1

Find du

Differentiate: du/dx = 2x, so du = 2x dx. That is exactly the rest of the integrand.

d u = 2 x d x

Rewrite in u and integrate

Replace x² + 1 with u and 2x dx with du. The power rule gives u⁶/6.

\int u^{5} d u = \frac{u^{6}}{6} + C

Substitute back

Put u = x² + 1 back so the answer is in terms of x.

\frac{{(x^{2} + 1)}^{6}}{6} + C

Check by differentiating

By the chain rule, the derivative of (x² + 1)⁶/6 is 6(x² + 1)⁵·2x/6 = 2x(x² + 1)⁵, the integrand. Expanding (x² + 1)⁵ first would also work, but it gives six terms and more room for slips.

Result

∫2x(x² + 1)⁵ dx = (x² + 1)⁶/6 + C.

Your turn

Find ∫x e^(x²) dx.

Show the answer and explanation

½e^(x²) + C.

Let u = x², so du = 2x dx and x dx = du/2. Then ∫eᵘ du/2 = eᵘ/2 + C = ½e^(x²) + C. Differentiating gives ½e^(x²)·2x = x e^(x²).

\begin{array}{l} F (x) = \frac{1}{2} e^{x^{2}} \\ F^{'} (x) = x e^{x^{2}} \end{array}

Keep exploring

In the checker, drop the 6 from the denominator. The derivative of (x² + 1)⁶ is 12x(x² + 1)⁵, six times the integrand, so the check fails.

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OpenStax Calculus Volume 1 — Substitution

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