How to integrate $\int \frac{(x^{4}+x^{7})^{\frac{1}{4}}}{x^{2}}dx$?

Question

How to evaluate $$\int \frac{(x^{4}+x^{7})^{\frac{1}{4}}}{x^{2}}dx$$ ? I have been trying this question for nearly one month but I can't solve it. I am trying this question by taking $x^{4}$ common. Now the integral will become $$\int \frac{x(1+x^{3})^{\frac{1}{4}}}{x^{2}}dx$$. Now I multiplied the numerator and denominator by $x$. (This was my trick). So, the integral will become $$\int \frac{x^{2}(1+x^{3})^{\frac{1}{4}}}{x^{3}}dx$$. Now, I substituted $x^{3}=u$. Then $x^{2}dx=\frac{1}{3}du$. So, the above integral will become $$\frac{1}{3}\int \frac{(1+u)^{\frac{1}{4}}}{u}du$$. Now I thought of substituting $(1+u)=t$. So, $du=dt$. Now, the above integral will become $$\frac{1}{3}\int \frac{t^{\frac{1}{4}}}{t-1}dt$$. Now I am thinking of substituting $t=\sec^{2}\theta$. Now $dt=2\sec^{2}\theta×\tan\theta×d\theta$.So, the above integral will become $$\frac{2}{3}\int \frac{(\sec\theta)^{\frac{5}{2}}}{\tan\theta}d\theta$$. But after this step, I can't approach further. Please help me out.

Answer 1

$$\int \frac{(x^{4}+x^{7})^{\frac{1}{4}}}{x^{2}}\mathrm dx=\int\frac{(1+x^3)^{1/4}}{x} \mathrm dx$$ An alternative to your method is to substitute $u=(1+x^3)^{1/4}$, $\mathrm du=\frac{3x^2}{4(1+x^3)^{3/4}}\mathrm dx$ to get: $$\int\frac{(1+x^3)^{1/4}}{x} \mathrm dx=\frac43\int \frac{u^4}{u^4-1}\mathrm du=\frac43\int\left(1+\frac{1}{u^4-1}\right)\mathrm du $$ and the last integral can be solved using partial fraction decomposition .

Answer 2

$$\int \frac{\sqrt[4]{x^4+x^7}}{x^2}\mathrm{d}x=\int \frac{x\sqrt[4]{1+x^3}}{x^2}\mathrm{d}x=\int \frac{\sqrt[4]{1+x^3}}{x}\mathrm{d}x$$ $$\int \frac{\sqrt[4]{1+x^3}}{x}\mathrm{d}x$$ $t=x^3$ $$\frac{1}{3}\int\frac{\sqrt[4]{1+t}}{t}\mathrm{d}t$$ $s=\sqrt[4]{t+1}$ $$\frac{4}{3}\int\frac{s^4}{s^3+1}\mathrm{d}s$$ Using this:

$$\frac{s^4}{s^3+1}=1-\frac{1}{2(s^2+1)}-\frac{1}{4(s+1)}+\frac{1}{4(s-1)}$$

$$\frac{4}{2}\cdot\frac{1}{6} \left(3 s^2 - \ln(s^2 - s + 1) + 2\ln(s + 1) - 2 \sqrt{3} \arctan\left(\frac{2 s - 1}{\sqrt3}\right)\right)$$

substituting the values ​​of $s$ into $t$ and from $t$ into $x$ you get your answer

Answer 3

$1+u=t^4$ was a good idea before you got stuck and losing your one month. Another late idea is to try $$t=\sec^4\theta.$$ Then you get $$\frac13\int\frac{t^{\frac14}}{t-1}dt=\frac43\int\frac{\sec^5\theta\tan\theta}{\sec^4\theta-1}d\theta$$ and by adding/substracting $\sec\theta\tan\theta$ in the numerator $$\frac43\int\left(1+\frac{\frac12}{\sec^2\theta-1}-\frac{\frac12}{\sec^2\theta+1}\right)\\\sec\theta\tan\theta\,d\theta \\ =\frac43\left(\sec\theta+\frac14\ln\left|\frac{\sec\theta-1}{\sec\theta+1}\right|-\frac12\arctan(\sec\theta)\right)$$ where $\sec\theta=(1+x^3)^{\frac14}.$

Answer 4

Here's a generalized approach (also known as Chebysev's Theorem)

$$I=\int x^m(a+bx^n)^p\,dx$$

Case (1):

If $p$ is an integer, substitute $x=t^N$, where $N$ is LCM of $m,n$.

Case (2):

If $\frac{m+1}{n}$ is an integer, substitute $a+bx^n=t^N$, where $N$ is denominator of $p$.

Case (3):

If $\frac{m+1}{n}+p$ is an integer, substitute $ax^{-1}+b=t^N$, where $N$ is denominator of $p$.

Given integral;

$$I=\int x^{-1}(1+x^3)^{\frac{1}{4}}\,dx$$

Above integral matches with case (2);

Substitute $1+x^3=t^4$ and proceed.