Mathematics

# Evaluate: $\displaystyle \int_{0}^{1} \dfrac {2x+3}{5x^2 +1}dx$

##### SOLUTION

$I=\displaystyle \int_0^1 \dfrac {2x+3}{5x^2 +1}dx$

$I=\displaystyle \int_0^1 \dfrac {2x}{5x^2 +1}dx +\displaystyle \int_0^1 \dfrac {3}{5x^2 +1}dx$

$I=\dfrac {1}{5} \displaystyle \int_0^1 \dfrac {10x}{5x^2 +1}dx +3\displaystyle \int_0^1 \dfrac {1}{(\sqrt {5} x)^2 +1} dx$

$\ I=\dfrac {1}{5}[\log (5x^2 +1)]_0^1 +\dfrac {3}{\sqrt 5}\left [\tan^{-1} \dfrac {\sqrt 5 x}{1}\right]_0^1$

$\ I=\dfrac {1}{5}(\log 6-\log 1)+\dfrac {3}{\sqrt 5}\tan^{-1}\sqrt 5$

$I=\dfrac {1}{5}\log 6+\dfrac {3}{\sqrt 5}\tan^{-1}\sqrt 5$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
if $\displaystyle \int \frac{x^{4} + 1}{x(x^{2} + 1)^{2}} dx = A\, ln |x| + \frac{B}{1 + x^{2}} + c$, where c is the constant of integration then :
• A. $A = 1, B = -1$
• B. $A = -1, B = 1$
• C. $A = -1, B = -1$
• D. $A = 1, B = 1$

1 Verified Answer | Published on 17th 09, 2020

Q2 Matrix Hard
The value of $\displaystyle \int_{a}^{b}f\left ( x \right )dx$
 $a= 0,\:b= \pi /3,\:f\left ( x \right )= \displaystyle \frac{x}{1+\sin x}$ $\displaystyle \frac{16}{3}\pi -2\sqrt{3}$ $a= 1,\:b= 16,\:f\left ( x \right )= \tan^{-1}\sqrt{\sqrt{x}-1}$ $\displaystyle \frac{\pi ^{2}}{16}+\displaystyle \frac{\pi }{4} \log 2$ $a= 0,\:b= \pi /2,\:f\left ( x \right )= \displaystyle \frac{x\sin 2x}{\left ( \sin x+\cos x \right )^{2}}$ $-\displaystyle \frac{\pi }{3} \left ( 2-\sqrt{3} \right )+\log \left ( \dfrac{1}{2}+\displaystyle \frac{\sqrt{3}}{2} \right )$ $a= 0,\:b= \pi /4,\:f\left ( x \right )= \displaystyle \frac{x^{2}\sec ^{2}x\left ( \cos 2x-\sin 2x \right )}{\left ( \sin x+\cos x \right )^{2}}$ $\displaystyle \frac{\pi ^{2}}{8}-\displaystyle \frac{\pi }{4}$

1 Verified Answer | Published on 17th 09, 2020

Q3 One Word Medium
$\int_0^8 {\frac{{\sqrt x }}{{\sqrt x + \sqrt {8 - x} }}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
By Simpson's rule, the value of $\displaystyle\int _{ -3 }^{ 3 }{ { x }^{ 4 }dy }$ by taking 6 sub-intervals, is
• A. $90$
• B. $80$
• C. $70$
• D. $98$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$