Question

Evaluate integral of $\int \frac{dx}{\sqrt[4]{41+x^4}}$, the ans is

• A. $-\frac{1}{4}[\log \frac{\sqrt[4]{41+1/x^4}-1}{\sqrt[4]{4(1+1/x^4)}+1}-2{\tan }^{-1}{(1+\frac{1}{x^4})}^{\frac{1}{4}}]$
• B. $-\frac{1}{4}[\log \frac{\sqrt[4]{41+1/x^4}-1}{\sqrt[4]{4(1+1/x^4)}+1}+2{\tan }^{-1}{(1+\frac{1}{x^4})}^{\frac{1}{4}}]+C$
• C. $\frac{1}{4}[\log \frac{\sqrt[4]{41+1/x^4}-1}{\sqrt[4]{4(1+1/x^4)}+1}+2{\tan }^{-1}{(1+\frac{1}{x^4})}^{\frac{1}{4}}]$
• D. $\frac{1}{4}[\log \frac{\sqrt[4]{41+1/x^4}-1}{\sqrt[4]{4(1+1/x^4)}+1}-2{\tan }^{-1}{(1+\frac{1}{x^4})}^{\frac{1}{4}}]+C$

Answer 1

The correct option is B $-\frac{1}{4}[\log \frac{\sqrt[4]{41+1/x^4}-1}{\sqrt[4]{4(1+1/x^4)}+1}+2{\tan }^{-1}{(1+\frac{1}{x^4})}^{\frac{1}{4}}]+C$

We write the given integral as $I=\int \frac{x^{4}dx}{x^5.\sqrt[4]{4(1+1/x^4)}},$
Substitute, $1+\frac{1}{x^4}=t^4$
So that $-\frac{4}{x^5}dx=4t^{3}dt$
$\therefore I=-\int \frac{t^3}{(t^4-1)t}dt=-\int \frac{t^{2}dt}{(t^2-1)(t^2+1)}=-\frac{1}{2}[\int \frac{dt}{t^2-1}+\int \frac{dt}{t^2+1}]$

$=-\frac{1}{2}[\frac{1}{2}\log \frac{t-1}{t+1}+{\tan }^{-1}t]=-\frac{1}{4}[\log \frac{\sqrt[4]{41+1/x^4}-1}{\sqrt[4]{4(1+1/x^4)}+1}+2{\tan }^{-1}{(1+\frac{1}{x^4})}^{\frac{1}{4}}]$