Solve indefinite integral $\int\frac{x^6}{\sqrt{x^2+1}}dx$

Question

I want to evaluate $$\int\frac{x^6}{\sqrt{x^2+1}}dx$$ I tried to use substitution $$x=tg(t)$$ and got $$\int\frac{tg(t)^6}{\cos(t)^3}dt$$

Answer 1

A hyperbolic trigonometric substitution gives a quick solution.

$x = \sinh y$

The integrand becomes simply $\sinh^6 y = \frac 1{64}(e^y - e^{-y})^6$, which is easily expanded with the binomial theorem and the termwise integration is trivial.

Answer 2

Your intuition is correct. Substitute $x = \tan(t)$. Then, $dx = \sec^2(t)dt$. This leads to $\sqrt{x^2 + 1}=\sqrt{\tan^2(t)+1} = \sec(t)$.

Substituting these terms yields $$ \int \frac{\tan^6(t)}{\sec(t)} \cdot \sec^2(t)dt = \int \tan^6(t)\sec(t) dt $$

From there you can apply $\tan^2(t) = \sec^2(t) -1$ and express the integrand using $\sec^2(t)$.

Answer 3

This type of integral is also known as a "Chebyshev" integral.

$$I = \int\frac{x^6}{\sqrt{x^2 + 1}}dx = \int x^6(x^2 + 1)^{1/2}dx$$

Where $m = 6, n = 2, p = \frac{1}{2}$. We observe that $\frac{m+1}{n} \notin \mathbb{Z}, p \notin \mathbb{Z}$, but $\frac{m+1}{n} + p = 4 \in \mathbb{Z}$. Therefore, consider $u = (1 + x^{-2})^{1/2}$. From then on you can express it in terms of rational functions.

Answer 4

• Euler-substitute:

$$\int \frac{x^6}{\sqrt{x^2+1}}\,dx \stackrel{x=\frac{2t}{1-t^2}}= 128 \int \frac{t^6}{(1-t^2)^7}\,dt$$

• Reduce the numerator to a cubic, then Euler-substitute:

$$\int \frac{x^6}{\sqrt{x^2+1}} \, dx \stackrel{t=x^2}= \int \frac{t^3}{2\sqrt{t^2+t}}\,dt \stackrel{t=\frac{u^2}{1-2u}}= \int \frac{u^6}{(1-2u)^4} \, du$$

The remaining integrands can be expanded into partial fractions.

Answer 5

Let $ I_n=\int \frac{x^n}{\sqrt{x^2+1}} dx$ and integrate both sides of $$\left(x^{n-1}\sqrt{x^2+1}\right)’ =\frac{n x^n}{\sqrt{x^2+1}}+ \frac{(n-1) x^{n-2}}{\sqrt{x^2+1}} $$ to establish the reduction formula $I_n = \frac{x^{n-1}}n\sqrt{x^2+1}-\frac{n-1}n I_{n-2} $. Then, apply it to $I_6$ three times $$\int \frac{x^6}{\sqrt{x^2+1}} dx =\bigg( \frac{x^5}6-\frac{5x^3}{24}+\frac{5x}{16}\bigg) \sqrt{x^2+1}-\frac5{16}I_0 $$ where $I_0= \int \frac{1}{\sqrt{x^2+1}} dx =\sinh^{-1}x$.