I=∫ x√(1+2x^2)dx Let 1+2x^2=t 4xdx=dt I=14∫√(t)dt I=14×23 t^3/2 +C =16 (1+2x^2)^3/2+C ...

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Integration by Substitution

Solve : ∫ x√...

Question

Solve :
$\int x \sqrt{1 + 2 x^{2}} d x$

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Solution

$I = \int x \sqrt{1 + 2 x^{2}} d x$
Let $1 + 2 x^{2} = t$
$4 x d x = d t$
$I = \frac{1}{4} \int \sqrt{t} d t$

$I = \frac{1}{4} \times \frac{2}{3} t^{3 / 2} + C$

$= \frac{1}{6} (1 + 2 x^{2})^{3 / 2} + C$

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