Mathematics

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

##### SOLUTION
$I = \displaystyle \int \frac{x^4 + 1}{x^6 + 1} dx = \displaystyle \int \frac{(x^2 + 1)^2 - 2x^2}{(x^2 + 1) (x^4 - x^2 + 1)} dx$
$= \displaystyle \int \frac{(1 + 1/x^2)dx}{(x^2 + 1/x^2 - 1)} - 2 \displaystyle \int \frac{x^2 dx}{(x^3)^2 + 1}$
$= \displaystyle \int \frac{(1 + 1/x^2) dx}{(x - 1/x)^2 + 1} - 2 \displaystyle \int \frac{x^2 dx}{(x^3)^2 + 1}$
In first integral put $x - 1/x = t$ and in second integral put $x^3 = u$
$= \displaystyle \int \frac{dt}{t^2 + 1} - \frac{3}{2} \displaystyle \int \frac{du}{u^2 + 1}$
$= \tan^{-1}(t) - \frac{2}{3} \tan^{-1} (u) + C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Subjective Medium
Solve $\displaystyle \int\limits_{\pi /6}^{\pi /3} {\frac{1}{{\sin 2x}}} \,dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int\limits_0^1 x \sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}dx}$
• A. $\frac{\pi }{4}$
• B. $\frac{1}{2}$
• C. $\frac{\pi }{4} + \frac{1}{2}$
• D. $\frac{\pi }{4} - \frac{1}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\displaystyle \int _{ 5 }^{ 10 }{ \left( \sqrt { x+\sqrt { 20x-100 } } -\sqrt { x-\sqrt { 20x-100 } } \right) }$ is equal to
• A. $5\sqrt {5}$
• B. $10\sqrt {2}$
• C. $5\sqrt {2}$
• D. $10\sqrt {5}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate the given integral.
$\displaystyle\int { \cfrac { 1 }{ { x }^{ 4 }-1 } } dx\quad$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$