# How do you find the integral of int x^3 * sqrt(x^2 + 4) dx?

May 29, 2018

$\frac{1}{15} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} \cdot \left(3 {x}^{2} - 8\right) + C$

#### Explanation:

$\int {x}^{3} \sqrt{{x}^{2} + 4} \cdot \mathrm{dx}$

After using $x = 2 \tan y$ and $\mathrm{dx} = 2 {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$ transforms, this integral became,

$\int {\left(2 \tan y\right)}^{3} \sqrt{{\left(2 \tan y\right)}^{2} + 4} \cdot 2 {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$

=$\int 16 {\left(\tan y\right)}^{3} \cdot {\left(\sec y\right)}^{2} \cdot \sqrt{4 {\left(\sec y\right)}^{2}} \cdot \mathrm{dy}$

=$\int 32 {\left(\sec y\right)}^{3} \cdot {\left(\tan y\right)}^{3} \cdot \mathrm{dy}$

=$\int 32 {\left(\sec y\right)}^{2} \cdot {\left(\tan y\right)}^{2} \cdot \sec y \cdot \tan y \cdot \mathrm{dy}$

=$\int 32 {\left(\sec y\right)}^{2} \cdot \left({\left(\sec y\right)}^{2} - 1\right) \cdot \sec y \cdot \tan y \cdot \mathrm{dy}$

After using $z = \sec y$ and $\mathrm{dz} = \sec y \cdot \tan y \cdot \mathrm{dy}$ transforms, it became

$\int 32 {z}^{2} \cdot \left({z}^{2} - 1\right) \cdot \mathrm{dz}$

=$\int \left(32 {z}^{4} - 32 {z}^{2}\right) \cdot \mathrm{dz}$

=$\frac{32}{5} {z}^{5} - \frac{32}{3} {z}^{3} + C$

=$\frac{32}{5} {\left(\sec y\right)}^{5} - \frac{32}{3} {\left(\sec y\right)}^{3} + C$

After using $x = 2 \tan y$, $\tan y = \frac{x}{2}$ and $\sec y = \frac{\sqrt{{x}^{2} + 4}}{2}$ inverse transforms, I found

$\frac{1}{5} {\left({x}^{2} + 4\right)}^{\frac{5}{2}} - \frac{4}{3} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} + C$

=$\frac{1}{15} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} \cdot \left(3 {x}^{2} - 8\right) + C$

May 29, 2018

$\frac{1}{15} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} \left(3 {x}^{2} - 8\right) + C$.

#### Explanation:

Let, $I = \int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = \int {x}^{2} \sqrt{{x}^{2} + 4} \cdot x \mathrm{dx}$.

Subst.

${x}^{2} + 4 = {t}^{2} , \mathmr{and} , {x}^{2} = {t}^{2} - 4. \therefore 2 x \mathrm{dx} = 2 t \mathrm{dt} , \mathmr{and} , x \mathrm{dx} = t \mathrm{dt}$

$\therefore I = \int \left({t}^{2} - 4\right) \cdot \sqrt{{t}^{2}} \cdot t \mathrm{dt} = \int \left({t}^{2} - 4\right) {t}^{2} \mathrm{dt}$,

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

$= {t}^{5} / 5 - 4 \cdot {t}^{3} / 3$,

$= {t}^{3} / 15 \left(3 {t}^{2} - 20\right)$,

$= \frac{1}{15} \cdot {t}^{2} \left(3 {t}^{2} - 20\right) t$.

Reverting from ${t}^{2} \to \left({x}^{2} + 4\right)$, we have,

$I = \frac{1}{15} \left({x}^{2} + 4\right) \left\{3 \left({x}^{2} + 4\right) - 20\right\} \sqrt{{x}^{2} + 4}$.

$\Rightarrow I = \frac{1}{15} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} \left(3 {x}^{2} - 8\right) + C$, as Respected Cem

Sentin has readily derived.

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