Mathematics

# Evaluate$\int {\dfrac{{dt}}{{5t + 1}}}$

##### SOLUTION
$\displaystyle\int\dfrac{dt}{5t+1}$
we know that $\displaystyle\int \dfrac{dx}{x}=\log 1\times 1+c$
put $u=5t+1$
$du=5dt$
$=\dfrac{1}{5}\displaystyle\int \dfrac{du}{u}$
$\dfrac{1}{5}\log |u|+C$
$\dfrac{1}{5}\log|5t+1|+C$

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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
$\displaystyle \int_{}^{} {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}} dx}$, where $0 < x < \dfrac{\pi }{2}$ is equal to
• A. $2{x^2} + C$
• B. ${x^2} + C$
• C. $\dfrac{x^3}{3} +C$
• D. $\dfrac{x^2}{2} + C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
lf $\displaystyle \int\frac{dx}{(\sin x+4)(\sin x-1)}=\frac{A}{\tan\frac{x}{2}-1}+B\tan^{-1}f(x)+c {\it}$ then
• A. $A =\displaystyle \frac{1}{5},\ B=\displaystyle \frac{2}{5\sqrt{15}},\ f(x) =\displaystyle \dfrac{4\tan x+3}{\sqrt{15}}$
• B. $A =-\displaystyle \frac{1}{5},\ B=\displaystyle \frac{2}{\sqrt{15}},\ f(x) =\displaystyle \frac{4\tan\dfrac{x}{2}+1}{\sqrt{15}}$
• C. $A=\displaystyle \frac{2}{5},\ B=-\displaystyle \dfrac{2}{5\sqrt{15}},f(x)=\frac{4\tan x+1}{\sqrt{15}}$
• D. $\displaystyle A=\frac{2}{5},\ B=-\displaystyle \frac{2}{5\sqrt{15}},f(x)=\frac{4\tan\dfrac{x}{2}+1}{\sqrt{15}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\int {\frac{{\sin x}}{{\sin \left( {x - \alpha } \right)}}dx = Ax + B\log \sin \left( {x - \alpha } \right) + C} ,$ then value of $(A,B)$ is
• A. $\left( {\sin \alpha ,\cos \alpha } \right)$
• B. $\left( { - \cos \alpha ,\sin \alpha } \right)$
• C. $\left( { - \sin \alpha ,\cos \alpha } \right)$
• D. $\left( {\cos \alpha ,\sin \alpha } \right)$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle \int \dfrac{\text{cosec}^2x - 2010}{\cos^{2010x}}dx=-\dfrac{f(x)}{(g(x))^{2010}}+C$; then the number of solutions where equation $\dfrac{f(x)}{g(x)}=\left \{ x \right \}$ in $[0, 2\pi]$ is/are:
• A. $1$
• B. $2$
• C. $3$
• D. $0$

$\int { \cfrac { f'(x) }{ f(x) } dx } =\log { [f(x)] } +c$