How do you find the antiderivative of int x(x^2+1)^100 dx?

Dec 31, 2016

$\left(\frac{1}{202}\right) {\left({x}^{2} + 1\right)}^{101} + C$

Explanation:

Write down ${\left({x}^{2} + 1\right)}^{101} + C$ as a guess and differentiate it (chain rule). You get $\left(101\right) . {\left({x}^{2} + 1\right)}^{101 - 1} . \left(2 x\right) = 202 {\left({x}^{2} + 1\right)}^{100}$. This is close, but too big by a factor of $202$. so divide the first guess by $202$.

Alternatively, substitute $u = {x}^{2} + 1$, giving $\frac{\mathrm{du}}{\mathrm{dx}} = 2 x$, $\frac{\mathrm{dx}}{\mathrm{du}} = \frac{1}{2 x}$.

Then by the substitution formula the integral becomes
$\int \cancel{x} . {u}^{100.} \left(\frac{1}{2 \cancel{x}}\right) \mathrm{du}$
$= \left(\frac{1}{2}\right) \int {u}^{100} \mathrm{du}$
$= \left(\frac{1}{2}\right) \left(\frac{1}{100 + 1}\right) {u}^{100 + 1} + C$ by the power law
$= \left(\frac{1}{202}\right) {\left({x}^{2} + 1\right)}^{101} + C$

Whatever you do, don't try to expand the bracket by the binomial expansion!