# How do you integrate (x^3)(x^2+1)^(1/2)?

Jun 24, 2016

${\left({x}^{2} + 1\right)}^{\frac{5}{2}} / 5 - {\left({x}^{2} + 1\right)}^{\frac{3}{2}} / 3 + C$

#### Explanation:

We have:

$\int {x}^{3} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} \mathrm{dx}$

We should try to use substitution. Let $u = {x}^{2} + 1$, implying that $\mathrm{du} = 2 x \mathrm{dx}$.

Note that we don't visibly have $2 x \mathrm{dx}$ in the integrand, but we can make some modifications.

$\int {x}^{3} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} \mathrm{dx} = \frac{1}{2} \int {x}^{3} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} 2 \mathrm{dx}$

Now, taking an $x$ from the ${x}^{3}$, or splitting it into ${x}^{3} = {x}^{2} \cdot x$:

$\frac{1}{2} \int {x}^{3} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} 2 \mathrm{dx} = \frac{1}{2} \int {x}^{2} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} 2 x \mathrm{dx}$

Here, we see that the leftover ${x}^{2}$ can be expressed in terms of $u$: ${x}^{2} = u - 1$.

$\frac{1}{2} \int {x}^{2} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} 2 x \mathrm{dx} = \frac{1}{2} \int \left(u - 1\right) {u}^{\frac{1}{2}} \mathrm{du}$

Now distributing the ${u}^{\frac{1}{2}}$:

$\frac{1}{2} \int \left(u - 1\right) {u}^{\frac{1}{2}} \mathrm{du} = \frac{1}{2} \int \left({u}^{\frac{3}{2}} - {u}^{\frac{1}{2}}\right) \mathrm{du}$

Splitting into two integrals and integrating with the rule $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$, where $n \ne - 1$:

$\frac{1}{2} \int \left({u}^{\frac{3}{2}} - {u}^{\frac{1}{2}}\right) \mathrm{du} = \frac{1}{2} \int {u}^{\frac{3}{2}} \mathrm{du} - \frac{1}{2} \int {u}^{\frac{1}{2}} \mathrm{du}$

$= \frac{1}{2} \left({u}^{\frac{5}{2}} / \left(\frac{5}{2}\right)\right) - \frac{1}{2} \left({u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right) + C = {u}^{\frac{5}{2}} / 5 - {u}^{\frac{3}{2}} / 3 + C$

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