Question

$\int \frac{x^4+1}{x^6+1}dx=$

• A. ${\tan }^{-1}x-{\tan }^{-1}{x}^3+c$
• B. ${\tan }^{-1}x-\frac{1}{3}{\tan }^{-1}\left(x^3\right)+c$
• C. ${\tan }^{-1}x+{\tan }^{-1}\left(x^3\right)+c$
• D. ${\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}\left(x^3\right)+c$

Answer 1

Answer: (D)

${\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}\left(x^3\right)+c$

$\int \frac{x^4+1}{x^6+1}dx$

$=\int \frac{x^4+1}{x^6+1}\frac{(x^2+1)}{(x^2+1)}dx$ [ multiplying numerator and denominator by $x^2+1$ ]

$=\int \frac{{x^6+x}^2+x^4+1}{(x^6+1)(x^2+1)}dx$

$=\int \frac{(x^6+1)+x^2(x^2+1)}{(x^6+1)(x^2+1)}dx$

$=\int \frac{(x^6+1)}{(x^6+1)(x^2+1)}dx+\int \frac{x^2(x^2+1)}{(x^6+1)(x^2+1)}dx$

$=\int \frac{dx}{x^2+1}+\frac{1}{3}\int \frac{3x^{2}dx}{(x^6+1)}$

$\Rightarrow x^3=t$ $\Rightarrow 3x^{2}dx=dt$

$=\int \frac{dx}{x^2+1}+\frac{1}{3}\int \frac{3x^{2}dx}{{\left(x^3\right)}^2+1}$

$=\int \frac{dx}{x^2+1}+\frac{1}{3}\int \frac{dt}{t^2+1}$

$={\tan }^{-1}x-1\frac{1}{3}{\tan }^{-1}t+c$

$={\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}\left(x^3\right)+c$

Hence, the answer is ${\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}\left(x^3\right)+c.$