U-substitution

If differentiation used the chain rule, you'd find an inside's derivative multiplying the result. U-substitution spots that pattern and undoes it.

Work through it

The inside is $x^{2} + 1$ , and its derivative is there. $u = x^{2} + 1, d u = 2 x d x$

Rewrite. $\int u^{3} d u$

Integrate. $\frac{u ^{4}}{4} + C$

Put $x$ back. $\frac{( x ^{2} + 1 ) ^{4}}{4} + C$

$u = 5 x$ , so. $d u = 5 d x$ , $d x = \frac{d u}{5}$

Rewrite. $\frac{1}{5} \int cos u d u$

Result. $\frac{1}{5} sin (5 x) + C$

$u = x^{2}$ , and limits change too. $u$ from 0 to 1

Rewrite. $\int_{0}^{1} e^{u} d u$

Evaluate. $e - 1$

Leaving an $x$ in the integral after substituting.

Everything must be in $u$ . If an $x$ remains, try a different $u$ .

The short version
Let

u

be the inside function. Replace

d x

using

d u

, and everything must end up in

u

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