Mathematics

# The value of the integers $\int\limits_0^2 {\dfrac{{\log ({x^2} + 2)}}{{{{\left( {x + 2} \right)}^2}}}dx\,is:}$

$\frac{{\sqrt 2 }}{3}\tan {\,^{ - 1}}\sqrt 2 + \frac{5}{{12}}\log \,2\, - \frac{1}{4}\log \,3$

##### SOLUTION
$I=\displaystyle\int^2_0\dfrac{log(x^2+2)}{(x+2)^2}$
Integrating by parts by $log(x^2+2)$ as first function and $\dfrac{1}{(x+2)^2}$ as second function
$\Rightarrow log(x^2+2)\displaystyle\int \dfrac{1}{(x+2)^2}dx+\displaystyle\int \dfrac{x}{x^2+2}\times \dfrac{1}{(x+2)}dx$
Now, $I_1=\displaystyle\int \dfrac{x}{(x^2+2)(x+2)}=A$
Let $\dfrac{x}{(x^2+2)(x+2)}=\dfrac{A}{x^2+2}+\dfrac{B}{x+2}$
$\dfrac{x}{(x^2+2)(x+2)}=\dfrac{Ax+2A+Bx^2+2B}{(x^2+2)(x+2)}$
On comparing we get $A=\dfrac{1}{3}, B=\dfrac{-1}{3}$
$\therefore I_1=\displaystyle\int \dfrac{x}{(x^2+2)(x+2)}=\displaystyle\int \left(\dfrac{1}{3(x^2+2)}-\dfrac{1}{3(x+2)}\right)dx$
$=\dfrac{2}{3}\tan^{-1}\dfrac{x}{\sqrt{2}}-\dfrac{1}{3}log|x+2|$
Now, $I=\left[\dfrac{-log (x^2+2)}{x+2}+\dfrac{2}{3}\tan^{-1}\dfrac{x}{\sqrt{2}}-log|x+2|\right]^2_0$
$=-\dfrac{log 6}{4}+\dfrac{2}{3}\tan^{-1}(\sqrt{2})-log 4+\dfrac{log 2}{2}-log 2$
$=\dfrac{z}{3}\tan^{-1}\sqrt{2}-\dfrac{5}{2}log^2-\dfrac{log 6}{4}$
$=\dfrac{2}{3}\tan^{-1}\sqrt{2}-\dfrac{5}{2}log^2-\dfrac{1}{4}[log^3+log^2]$
$=\dfrac{2}{3}\tan^{-1}\sqrt{2}-\dfrac{1}{4}log^3-\dfrac{11}{2}log^2$.

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

#### Realted Questions

Q1 Single Correct Medium
Evaluate: $\displaystyle \int_{0}^{\pi /2} \sin^{3}x.\cos^{3}x dx$
• A. $\displaystyle \frac{\pi}{24}$
• B. $\displaystyle \frac{\pi}{12}$
• C. $\displaystyle \frac{1}{24}$
• D. $\displaystyle \frac{1}{12}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Using properties of definite integrals, evaluate $\displaystyle \int_{0}^{\tfrac{\pi}2} \dfrac {\sqrt {\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The value of $2\displaystyle \int \sin x \text{ cosec } 4x\ dx$ is equal to:
• A. $\dfrac{1}{2\sqrt2}\ln \left|\dfrac{1+\sqrt2\sin x}{1-\sqrt2\sin x}\right|- \dfrac{1}{2}\ln \left|\dfrac{1+\sin x}{\cos x}\right|+c$
• B. $\dfrac{1}{2\sqrt2}\ln \left|\dfrac{1-\sqrt2\sin x}{1+\sqrt2\sin x}\right|- \dfrac{1}{4}\ln \left|\dfrac{1+\sin x}{1-\sin x}\right|+c$
• C. $\dfrac{1}{2\sqrt2}\ln \left|\dfrac{1-\sqrt2\sin x}{1+\sqrt2\sin x}\right|+ \dfrac{1}{2}\ln \left|\dfrac{1+\sin x}{\cos x}\right|+c$
• D. $\dfrac{1}{2\sqrt2}\ln \left|\dfrac{1+\sqrt2\sin x}{1-\sqrt2\sin x}\right|- \dfrac{1}{4}\ln \left|\dfrac{1+\sin x}{1-\sin x}\right|+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate $\int \dfrac{1}{x^2+2x+2} dx$
• A. $\tan^{-1}(x+2)+C$
• B. $\tan^{-1}(x+3)+C$
• C. None of these
• D. $\tan^{-1}(x+1)+C$

Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively.