Mathematics

# If $\displaystyle \frac{ax+b}{(3x+4)^2} =\frac{1}{3x+4}-\frac{3}{(3x+4)^2}$ then

$b=1$

$a=3$

##### SOLUTION
Given, $\displaystyle \frac{ax+b}{(3x+4)^2} =\frac{1}{3x+4}-\frac{3}{(3x+4)^2}$       ....(1)
Resolving into partial fractions
$\displaystyle \frac { ax+b }{ (3x+4)^{ 2 } } =\frac { A }{ 3x+4 } +\frac { B }{ (3x+4)^{ 2 } }$      .....(2)
$\displaystyle \frac { ax+b }{ (3x+4)^{ 2 } } =\frac { A(3x+4)+B }{ (3x+4)^{ 2 } }$
$\Rightarrow ax+b=A(3x+4)+B$
$\Rightarrow a=3A,b=4A+B$
$\Rightarrow \displaystyle A=\frac{a}{3}, B=b-\frac{4a}{3}$
Put these values in (2)
$\displaystyle \frac { ax+b }{ (3x+4)^{ 2 } } =\frac { a }{ 3(3x+4) } +\frac { 3b-4a }{ 3(3x+4)^{ 2 } }$
Comparing this with (1),
$\displaystyle \frac{a}{3}=1,\frac { 3b-4a }{ 3}=-3$
$\Rightarrow a=3,b=1$

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int e^{-x}[\frac{2x+5}{(2x+3)^{2}}]dx=$
• A. $\displaystyle \frac{e^{-x}}{2x+3}+c$
• B. $e^{-x}\displaystyle \frac{1}{2x+5}+c$
• C. $-e^{-x}\displaystyle \frac{1}{2x+5}+c$
• D. $\displaystyle \frac{-e^{-x}}{2x+3}+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Find the following integrals
$\displaystyle\int \cfrac{x^{3}-x^{2}+x-1}{x-1} d x$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle \int \frac{2\sin x}{(3+\sin 2x)}dx$ is equal to
• A. $\displaystyle \frac{1}{2}\ln \left | \frac{2+\sin x-\cos x}{2-\sin x+\cos x} \right |-\frac{1}{\sqrt{2}}\tan ^{-1}x\left ( \frac{\sin x+\cos x}{\sqrt{2}} \right )+c$
• B. $\displaystyle \frac{1}{2}\ln \left | \frac{2+\sin x-\cos x}{2-\sin x+\cos x} \right |-\frac{1}{2\sqrt{2}}\tan ^{-1}x\left ( \frac{\sin x+\cos x}{\sqrt{2}} \right )+c$
• C. none of these
• D. $\displaystyle \frac{1}{4}\ln \left | \frac{2+\sin x-\cos x}{2-\sin x+\cos x} \right |-\frac{1}{\sqrt{2}}\tan ^{-1}x\left ( \frac{\sin x+\cos x}{\sqrt{2}} \right )+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Solve $\int {\cos 2x\cos 4xcos6xdx}$.
• A. $I= {\dfrac{1}{4}[\dfrac{\sin12x}{12}+\dfrac{\sin8x}{8}-\dfrac{\sin4x}{4}+x]+c}$
• B. $I= {\dfrac{5}{4}[\dfrac{\sin12x}{12}-\dfrac{\sin8x}{8}+\dfrac{\sin4x}{4}+x]+c}$
• C. None of these
• D. $I= {\dfrac{1}{4}[\dfrac{\sin12x}{12}+\dfrac{\sin8x}{8}+\dfrac{\sin4x}{4}+x]+c}$

Evaluate $\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx$