evaluate $\int \frac{x^3}{\sqrt{x^2+1}}dx$

Question

Need help evaluating

$\int \frac{x^3}{\sqrt{x^2+1}}dx$

I want to know why my working out is illegitimate. It looks illegal. Im just not sure by what mechanism it is illegal.

Answer 1

Here is another way to approach it using hyperbolic functions: \begin{align*} \int\frac{x^{3}}{\sqrt{x^{2} + 1}}\mathrm{d}x & = \int\frac{\sinh^{3}(u)}{\sqrt{\sinh^{2}(u) + 1}}\mathrm{d}(\sinh(u))\\ & = \int\sinh^{3}(u)\mathrm{d}u\\ & = \int\sinh^{2}(u)\sinh(u)\mathrm{d}u\\ & = \int(\cosh^{2}(u) - 1)\sinh(u)\mathrm{d}u\\ & = \int(\cosh^{2}(u) - 1)\mathrm{d}(\cosh(u))\\ & = \int(v^{2} - 1)\mathrm{d}v\\ & = \frac{v^{3}}{3} - v + C\\ & = \frac{\cosh^{3}(u)}{3} - \cosh(u) + C\\ & = \frac{(x^{2} + 1)^{3/2}}{3} - \sqrt{x^{2} + 1} + C \end{align*}

Hopefully this helps!

Answer 2

$\int \frac{x^3}{\sqrt{x^2+1}}dx$

let $x=\tan(u)$

so $dx=\sec^2(u)du$

so the integral is now $$ \int \frac{\tan^3(u)(\sec^2(u))du}{\sec(u)} $$

which is the same as $$ \int \tan^3(u)\sec(u)du$$ or $$\int ((\sec^2(u)-1)(\sec(u)\tan(u))du$$

now sub $\sec(u)=t$ and you should get your answer. Please ping me If I've made any errors

Answer 3

The penultimate equality is wrong. Instead, in $$\int \sec \theta \tan^3 \theta \,d\theta $$ the powers of both $\tan \theta$ and $\sec \theta$ are odd, which suggests the substitution $v = \sec \theta, dv = \sec \theta \tan \theta \,d\theta,$ giving $$\int \sec \theta \tan^3 \theta \,d\theta = \int (\sec^2 \theta - 1) \sec \theta \tan \theta \,d\theta = \int (v^2 - 1) \,dv = \frac{1}{3} v^3 - v + C .$$

Composing the substitutions $x \rightsquigarrow \theta \rightsquigarrow v$ gives another possible choice, $v = \sqrt{x^2 + 1}, dv = \frac{x \,dx}{\sqrt{x^2 + 1}}$.

Remark There's an arguably easier method to evaluate the original integral that also avoids trigonometric/hyperbolic substitutions. Substituting $w = x^2 + 1$, $dw = 2 x \,dx$, gives $$\int \frac{x^3 \,dx}{\sqrt{x^2 + 1}} = \frac12 \int \frac{(w - 1) \,dw}{\sqrt w} = \frac12 \int \left(\sqrt w - \frac1{\sqrt w}\right) = \frac13 w^{3/2} - \sqrt w + C .$$

More generally, given an integral of the form $$\int x^{2 m + 1} f(x^2 + a) \,dx,$$ where $m$ is an integer, the substitution $w = x^2 + a$, $dw = 2 x \,dx$ transforms the integral to $$\frac12 \int (w - a)^m f(w) \,dw .$$

Answer 4

Integration by parts $$ \begin{aligned} \int \frac{x^3}{\sqrt{x^2+1}}\, d x&=\int x^2 d\left(\sqrt{x^2+1}\right) \\& =x^2 \sqrt{x^2+1}-\int \sqrt{x^2+1} \,d\left(x^2+1\right) \\ & =x^2 \sqrt{x^2+1}-\frac 32\left(x^2+1\right)^{\frac{1}{2}}+C\\ & =\frac{\sqrt{x^2+1}}{3}\left(x^2-2\right)+C \end{aligned} $$