Question

$\int e^{3\log x}{(x^4+1)}^{-1}dx$ is equal to ?

• A. $e^{3\log x}$
• B. $\frac{1}{4}\log (x^4+1)+C$
• C. $\log (x^4+1)+C$
• D. $\frac{1}{2}\log (x^4+1)+C$

Answer 1

The correct option is B $\frac{1}{4}\log (x^4+1)+C$

$\int e^{3\log x}(x^4+1{)}^{-1}dx=\int e^{\log x^3}(x^2+1{)}^{-1}dx$

$=\int \frac{x^3}{1+x^4}dx$

Let $14x^4=t$

$x^{3}dx=dt$

$\Rightarrow =\int \frac{dt}{4t}$

$=\frac{1}{4}\log t+c$

$=\frac{1}{4}\log (1+x^4)+c$