Mathematics

# Evaluate $\int _ { 0 } ^ { 2 } \frac { 6 x + 3 } { x ^ { 2 } + 4 } d x$

##### SOLUTION

$\\\int_{0}^{2}(\frac{6x+3}{x^2+4})dx\\=3\int_{0}^{2}(\frac{2x}{x^2+4})dx+3\int_{0}^{2}(\frac{1}{x^2+2^2})dx\\=3[log(x^2+4)]_{0}^{2}+3\times(\frac{1}{2})\left[tan^{-1}(\frac{x}{2})\right]_{0}^{2}\\=3(log8-log4)+(\frac{3}{2})(tan^{-1}1-tan^{-1}0)\\=3log2+(\frac{3}{2})\times(\frac{\pi}{4})\\=3log2+(\frac{3\pi}{8})$

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int \frac {dx}{e^x+e^{-x}}$ is equal to
• A. $\tan^{-1}(e^{-x})+C$
• B. $\log (e^x-e^{-x})+C$
• C. $\log (e^x+e^{-x})+C$
• D. $\tan^{-1}(e^x)+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int x^{3}e^{x^{2}}dx=$
• A. $e^{x^{2}}(x^{2}-1)+c$
• B. $\dfrac{1}{2}e^{x^{2}}(x^{2}+1)+c$
• C. $e^{x^{2}}(x^{2}+1)+c$
• D. $\displaystyle \frac{1}{2}e^{x^{2}}(x^{2}-1)+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle f(x)=\lim_{n\rightarrow \infty }(2x+4x^{3}+......+2^{n}x^{2n-1})\left ( 0<x<\frac{1}{\sqrt{2}} \right )$, then the value of $\displaystyle\int f(x) dx$ is equal to
$\textbf{Note}$: $c$ is the constant of integration.
• A. $\displaystyle \log\left ( \frac{1}{\sqrt{1-x^{2}}} \right )+c$
• B. $\displaystyle \log\sqrt{1-2x^{2}+x} + c$
• C. None of these
• D. $\displaystyle \log\left ( \frac{1}{\sqrt{1-2x^{2}}} \right )+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following integral
$\int { \cfrac { -\sin { x } +2\cos { x } }{ 2\sin { x } +\cos { x } } } dx$

$\int \frac{1}{1+x}\;dx$