Question

$\int \frac{5}{x^4+1}dx$

• A.
• B.
• C.
• D. .

Answer 1

The correct option is B

We can see that the given form can be converted into one of the algebraic twins form in which we have “2” in the numerator.

So, to do this we’ll write it like -

$\frac{5}{2}\int \frac{2}{x^4+1}dx$

Let’s call this integral $\int \frac{2}{x^4+1}dxasI$

$I=\int \frac{2}{x^4+1}dx$

Now we can see that is in algebraic twins form. So, to solve it, we’ll apply the same approach which we know for such forms

$I=\int \frac{2}{x^4+1}dx$ can be written as -

$I=\int \frac{1+x^2}{x^4+1}dx+\int \frac{1-x^2}{x^4+1}dx$

$I=\int \frac{1+x^2}{x^4+1}dx-\int \frac{x^2-1}{x^4+1}dx$

Let’s divide $x^2$in the numerator or denominator -

$I=\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx-\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$

$OrI=\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+2}dx-\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-2}dx$

Let’s deal with both the integrals separately.

Let’s call first integral as $I_1$

$So,I_1=\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+2}dx$

$Substitutex-\frac{1}{x}=t$

$\&(1+\frac{1}{x^2}).dx=dt$

$I_1=\int \frac{1}{(t{)}^2+{\left(\sqrt{2}\right)}^2}dt$

Using the standard formulae we can say

$I_1=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{t}{2}\right)+C_1$

$Or{I}_1=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C_1$

Let’s call the second integral as

$I_2=\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-2}$

$Substitudex+\frac{1}{x}=u$

$\&(1-\frac{1}{x^2})dx=du$

$I_2=\int \frac{1}{(u{)}^2-(\sqrt{2}{)}^2}du$

Using the standard formulae we can say

$I_2=\frac{1}{2\sqrt{2}}log\left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+C_2$

So, the final answer will be -

$=\frac{5}{2}[\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{2\sqrt{2}}log\left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|]+c$