# What is the antiderivative of (x^3)sqrt(x^2 + 1)) dx?

Apr 13, 2015

The answer is: $\frac{1}{15} \left({x}^{2} + 1\right) \sqrt{{x}^{2} + 1} \left(3 {x}^{2} - 2\right) + c$.

If we make the substitution:

$x = \sinh t \Rightarrow \mathrm{dx} = \cosh t \mathrm{dt}$, than:

$\int {x}^{3} \sqrt{{x}^{2} + 1} \mathrm{dx} = \int {\sinh}^{3} t \sqrt{{\sinh}^{2} t + 1} \cdot \cosh t \mathrm{dt} =$

$= \int {\sinh}^{3} t \sqrt{{\cosh}^{2} t} \cdot \cosh t \mathrm{dt} = \int {\sinh}^{3} t {\cosh}^{2} t \mathrm{dt} =$

$= \int {\sinh}^{2} t \sinh t {\cosh}^{2} t \mathrm{dt} = \int \left({\cosh}^{2} t - 1\right) \sinh t {\cosh}^{2} t \mathrm{dt} =$

$= \int \left(\sinh t {\cosh}^{4} t - \sinh t {\cosh}^{2} t\right) \mathrm{dt} =$

$= {\cosh}^{5} \frac{t}{5} - {\cosh}^{3} \frac{t}{3} + c = \left(1\right)$

And, since $\cosh t = \sqrt{{\sinh}^{2} t + 1} = \sqrt{{x}^{2} + 1}$,

$\left(1\right) = {\left(\sqrt{{x}^{2} + 1}\right)}^{5} / 5 - {\left(\sqrt{{x}^{2} + 1}\right)}^{3} / 3 + c =$

$= {\left({x}^{2} + 1\right)}^{2} \frac{\sqrt{{x}^{2} + 1}}{5} - \left({x}^{2} + 1\right) \frac{\sqrt{{x}^{2} + 1}}{3} + c =$

$= \left({x}^{2} + 1\right) \sqrt{{x}^{2} + 1} \left[\frac{{x}^{2} + 1}{5} - \frac{1}{3}\right] + c =$

$= \left({x}^{2} + 1\right) \sqrt{{x}^{2} + 1} \left(\frac{3 {x}^{2} + 3 - 5}{15}\right) + c =$

$= \frac{1}{15} \left({x}^{2} + 1\right) \sqrt{{x}^{2} + 1} \left(3 {x}^{2} - 2\right) + c$.