# Evaluate the following integrals or show that the integral diverges. \int_{-\infty}^{\infty}\frac{x^{3}}{1+x^{8}}dx

Evaluate the following integrals or show that the integral diverges.
$$\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{x}^{{{8}}}}}}{\left.{d}{x}\right.}$$

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nghodlokl

Step 1
Given- $$\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{x}^{{{8}}}}}}{\left.{d}{x}\right.}$$
To evaluate- The above improper integral.
Concept Used- The above integral can be solved by using the substitution method.
Step 2
Explanation- Rewrite the given integral as,
$$\displaystyle{I}={\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{x}^{{{8}}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{\left({x}^{{{4}}}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}$$
Now, substituting $$x^{4}=t$$ and differentiating w.r.t. x, we get,
$$\displaystyle{4}{x}^{{{3}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}$$
$$\displaystyle{x}^{{{3}}}{\left.{d}{x}\right.}={\frac{{{\left.{d}{t}\right.}}}{{{4}}}}$$
At $$\displaystyle{x}=-\infty,{t}=-\infty$$ and at $$\displaystyle{x}=\infty,{t}=\infty$$.
Now, from the above integral,we can write as,
Step 3
$$\displaystyle={\infty_{{-\infty}}^{{\infty}}}{\frac{{{\frac{{{\left.{d}{t}\right.}}}{{{4}}}}}}{{{1}+{t}^{{{2}}}}}}$$
$$\displaystyle={\frac{{{1}}}{{{4}}}}{\int_{{-\infty}}^{{\infty}}}{\frac{{{1}}}{{{1}+{t}^{{{2}}}}}}{\left.{d}{t}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{4}}}}{{\left[{{\tan}^{{-{1}}}{\left({t}\right)}}\right]}_{{-\infty}}^{{+\infty}}}$$
$$\displaystyle={\frac{{{1}}}{{{4}}}}{\left[{{\tan}^{{-{1}}}{\left(\infty\right)}}-{{\tan}^{{-{1}}}{\left(-\infty\right)}}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{4}}}}{\left[{\frac{{\pi}}{{{2}}}}-{\left({\frac{{-\pi}}{{{2}}}}\right)}\right]}$$
$$\displaystyle={\frac{{\pi}}{{{4}}}}$$
Answer- Hence, the value of the integral $$\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{x}^{{{3}}}}}{{{1}+{x}^{{{8}}}}}}{\left.{d}{x}\right.}$$ is $$\displaystyle{\frac{{\pi}}{{{4}}}}$$.

###### Not exactly what you’re looking for?
Marcus Herman
It is required to calculate:
$$\displaystyle\int{\frac{{{x}^{{{3}}}}}{{{x}^{{{8}}}+{1}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{4}}}}\int{\frac{{{1}}}{{{u}^{{{2}}}+{1}}}}{d}{u}$$
Now we calculate:
$$\displaystyle\int{\frac{{{1}}}{{{u}^{{{2}}}+{1}}}}{d}{u}$$
This is the well-known tabular integral:
$$\displaystyle={\arctan{{\left({u}\right)}}}$$
We substitute the already calculated integrals:
$$\displaystyle{\frac{{{1}}}{{{4}}}}\int{\frac{{{1}}}{{{u}^{{{2}}}+{1}}}}{d}{u}$$
$$\displaystyle={\frac{{{\arctan{{\left({u}\right)}}}}}{{{4}}}}$$
Reverse replacement $$\displaystyle{u}={x}^{{{4}}}:$$
$$\displaystyle={\frac{{{\arctan{{\left({x}^{{{4}}}\right)}}}}}{{{4}}}}$$
$$\displaystyle\int{\frac{{{x}^{{{3}}}}}{{{x}^{{{8}}}+{1}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{\arctan{{\left({x}^{{{4}}}\right)}}}}}{{{4}}}}+{C}$$
nick1337

We put the expression $$4*x^{3}$$ under the differential sign, i.e.:
$$4x^{3}dx=d(x^{4}), t=x^{4}$$
Then the original integral can be written as follows:
$$\int \frac{1}{4*(t^{2}+1)}*dt$$
This is a table integral:
$$\int \frac{1}{4*(t^{2}+1)}*dt=\frac{\arctan (t)}{4}+C$$
To write down the final answer, it remains replace t with $$x^{4}$$
$$\frac{\arctan (x^{4})}{4}+C$$