Mathematics

# Evaluate $i)\displaystyle \int \dfrac{x^2 + 1}{x^4 + 1}$ dx$ii)\displaystyle \int \dfrac{dx}{x^2 + 1}$

##### SOLUTION
$i)\displaystyle \int{\cfrac{{x}^{2} + 1}{{x}^{4} + 1} dx}$
$= \displaystyle\int{\cfrac{1 + \cfrac{1}{{x}^{2}}}{{x}^{2} + \cfrac{1}{{x}^{2}}} dx}$
$= \displaystyle \int{\cfrac{1 + \cfrac{1}{{x}^{2}}}{{\left( x - \cfrac{1}{x} \right)}^{2} + 2} dx}$
Let
$x - \cfrac{1}{x} = t \Rightarrow \left( 1 + \cfrac{1}{{x}^{2}} \right) dx = dt$
Therefore, the integral  become-
$\int{\cfrac{dt}{{t}^{2} + {\left( \sqrt{2} \right)}^{2}}}$
$= \cfrac{1}{\sqrt{2}} \tan^{-1}{\left( \cfrac{t}{\sqrt{2}} \right)} + C$
Substituting the value of $t$, we get
$= \cfrac{1}{\sqrt{2}} \tan^{-1}{\left( \cfrac{\left( 1 + \cfrac{1}{{x}^{2}} \right)}{\sqrt{2}} \right)} + C$
$= \cfrac{1}{\sqrt{2}} \tan^{-1}{\left( \cfrac{{x}^{2} + 1}{\sqrt{2} {x}^{2}} \right)} + C$

$ii)\displaystyle \int{\cfrac{1}{{x}^{2} + 1} dx}$
$\displaystyle \int{\cfrac{1}{{x}^{2} + {\left( 1 \right)}^{2}} dx}$
$= \tan^{-1}{x} + C$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle\int^{\pi/2}_0\dfrac{\sin^nx}{(\sin^nx+\cos^nx)}dx=?$
• A. $\dfrac{\pi}{2}$
• B. $1$
• C. $0$
• D. $\dfrac{\pi}{4}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int\dfrac{(\cos x)^{n-1}}{(\sin x)^{n+1}}dx=$
• A. $\displaystyle \frac{-\cot^{n}x}{n+1}+c$
• B. $\displaystyle \frac{\cot^{n}x}{n}+c$
• C. $\displaystyle \frac{\cot^{n}x}{n+1}+c$
• D. $\displaystyle \frac{-\cot^{n}x}{n}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Statement-l $\displaystyle \int_{0}^{\pi/2}\frac{dx}{1+\tan^{5}x}=\frac{\pi}{4}$

Statement 2:$\displaystyle \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a+x)dx= \int_{0}^{\pi/2}\displaystyle \frac{dx}{1+\tan^{3}x}=\int_{0}^{\pi/2}\displaystyle \frac{d_{X}}{1+\cot^{3}x}=\frac{\pi}{4}$
• A. Statement 1 is True, Statement 2 is True; Statement 2 is a correct exlanation for Statement 1
• B. Statement 1 is True, Statement 2 is True; Statement 2 Not a correct exlanation for Statement 1
• C. Statement 1 is False, Statement 2 is True
• D. Statement 1 is True, Statement 2 is False

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int \dfrac{dt}{t + \sqrt{a^2 - t^2}}$ equal to
• A. $\dfrac{1}{2} \sin^{-1} \left(\dfrac{t}{a} \right) + \log (t + \sqrt{a^2 - t^2}) + k$
• B. $\dfrac{1}{2} \sin^{-1} \left(\dfrac{t}{a} \right) + \log \sqrt{a + \sqrt{a^2 - t^2}} + k$
• C. $\dfrac{1}{2} \sin^{-1} \left(\dfrac{t}{a} \right) + \log (a + \sqrt{a^2 - t^2}) + k$
• D. $\dfrac{1}{2} \sin^{-1} \left(\dfrac{t}{a} \right) + \log \sqrt{t + \sqrt{a^2 - t^2}} + k$

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 $\displaystyle f(x)>0 \forall x \in R$