Mathematics

# Solve :$\int_{}^{} {\frac{{\left( {3x - 2} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x + 3} \right)}}} dx$

##### SOLUTION
$\begin{array}{l} \int _{ }^{ }{ \frac { { \left( { 3x-2 } \right) } }{ { { { \left( { x+1 } \right) }^{ 2 } }\left( { x+3 } \right) } } } dx \\ Let\, \frac { { \left( { 3x-2 } \right) } }{ { { { \left( { x+1 } \right) }^{ 2 } }\left( { x+3 } \right) } } =\frac { A }{ { \left( { x+1 } \right) } } +\frac { B }{ { { { \left( { x+1 } \right) }^{ 2 } } } } +\frac { C }{ { \left( { x+3 } \right) } } \\ \Rightarrow \left( { 3x-2 } \right) \equiv A\left( { x+1 } \right) \left( { x+3 } \right) +B\left( { x+3 } \right) +C{ \left( { x+1 } \right) ^{ 2 } }.......\left( i \right) \\ Putting\, \, x=-13\, no\, both\, side\, of\, \left( i \right) ,\, we\, get\, C=\frac { { -11 } }{ 4 } \\ Putting\, x=-1\, on\, both\, sides\, of\left( i \right) ,\, we\, get\, B\, =\frac { { -5 } }{ 2 } \\ A+C=0\Rightarrow A=-C=\frac { { 11 } }{ 4 } \\ \therefore \frac { { \left( { 3x-2 } \right) } }{ { { { \left( { x+1 } \right) }^{ 2 } }\left( { x+3 } \right) } } =\frac { { 11 } }{ { 4\left( { x+1 } \right) } } -\frac { 5 }{ { 2{ { \left( { x+1 } \right) }^{ 2 } } } } -\frac { { 11 } }{ { 4\left( { x+3 } \right) } } \\ \Rightarrow \int _{ }^{ }{ \frac { { \left( { 3x-2 } \right) } }{ { { { \left( { x+1 } \right) }^{ 2 } }\left( { x+3 } \right) } } } dx \\ =\frac { { 11 } }{ 4 } \int _{ }^{ }{ \frac { { dx } }{ { \left( { x+1 } \right) } } } -\frac { 5 }{ 2 } \int _{ }^{ }{ \frac { 1 }{ { { { \left( { x+1 } \right) }^{ 2 } } } } } dx-\frac { { 11 } }{ 4 } \int _{ }^{ }{ \frac { { dx } }{ { \left( { x+3 } \right) } } } \\ =\frac { { 11 } }{ 4 } .\log \left| { \frac { { x+1 } }{ { x+3 } } } \right| +\frac { 5 }{ { 2\left( { x+1 } \right) } } +C \end{array}$

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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
$\int \sin x . \cos x . \cos 2x . \cos 4x . \cos 8x . \cos 16x\ dx$ equals.
• A. $\dfrac {\sin 16x}{1024} + c$
• B. $\dfrac {\cos 32x}{1096} + c$
• C. $-\dfrac {\cos 32x}{1096} + c$
• D. $\dfrac {-\cos 32x}{1024} + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Let $f(x) = e^{sinx}$ and $g(2-x) = x^{2} - f(x)$.
On the basis of above information, answer the following question :
If $\displaystyle \int_{0}^{2}(f(x)+g(x))dx=k$ then $[k]$ is $($where $[.]$ denotes greatest integer function$)$
• A. $0$
• B. $4$
• C. $5$
• D. $2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\int e^x\left(log x+\frac{1}{x^2}\right)dx=$
• A. $e^x\left(lnx+\frac{1}{x}\right)+C$
• B. $e^x\left(-lnx-\frac{1}{x}\right)+C$
• C. $e^x\left(-lnx+\frac{1}{x}\right)+C$
• D. $e^x[logx-(\frac{1}{x})]+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate $\displaystyle\int_{1}^{2}\dfrac{1}{x(1+\log x)^{2}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$