Mathematics

# Partial fraction from of $\displaystyle \frac{3x+7}{x^2-3x+2}$ is

$\displaystyle \frac{13}{x-2}-\frac{10}{x-1}$

##### SOLUTION
Let $\displaystyle \frac { 3x+7 }{ x^{ 2 }-3x+2 } =\frac { A }{ \left( x-2 \right) } +\frac { B }{ \left( x-1 \right) }$
$\Rightarrow 3x+7=A\left( x-1 \right) +B\left( x-2 \right)$
On comparing coefficients $3=A+B,7=-A-2B\Rightarrow A=13,B=-10$
Thus
$\displaystyle \frac { 3x+7 }{ x^{ 2 }-3x+2 } =\frac { 13 }{ \left( x-2 \right) } -\frac { 10 }{ \left( x-1 \right) }$
Hence, option 'A' is correct.

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

#### Realted Questions

Q1 One Word Medium
$\displaystyle \int_{0}^{1}x\left ( 1-x \right )^{4}dx= \frac{1}{C}$, then $C=?$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle I= \int\dfrac{\ln \left ( \dfrac{x-1}{x+1} \right )}{x^{2}-1}dx$ is equal to
• A. $\displaystyle \dfrac{1}{2}\left ( \ln \left ( \dfrac{x-1}{x+1} \right ) \right )^{2}+C$
• B. $\displaystyle \dfrac{1}{2}\left ( \ln \left ( \dfrac{x+1}{x-1} \right ) \right )^{2}+C$
• C. $\displaystyle \dfrac{1}{4}\left ( \ln \left ( \dfrac{x+1}{x-1} \right ) \right )$
• D. $\displaystyle \dfrac{1}{4}\left ( \ln \left ( \dfrac{x-1}{x+1} \right ) \right )^{2}+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate: $\int_{0}^{\pi}\dfrac {4x \sin x}{1 + \cos^{2}x} dx.$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\int\dfrac{dx}{\sqrt{1-x^2}-1}=$
• A. $\dfrac{1+\sqrt{1-x^2}}{x}-2tan^{-1}\sqrt{\dfrac{1+x}{1-x}}+C$
• B. $\dfrac{1+\sqrt{1-x^2}}{x}+2tan^{-1}\sqrt{\dfrac{1+x}{1-x}}+C$
• C. $\dfrac{1+\sqrt{1-x^2}}{x}+2tan^{-1}\sqrt{\dfrac{1-x}{1+x}}+C$
• D. $\dfrac{1+\sqrt{1-x^2}}{x}-2tan^{-1}\sqrt{\dfrac{1-x}{1+x}}+C$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$