Mathematics

# $\displaystyle \int \dfrac{x+3}{(x+1)^{4}}dx$

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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
If $\int f ( x ) d x = f ( x ) ,$ then $\int [ j ( x ) ] ^ { 2 } d x$ is

• A. $[ f ( x ) ] ^ { 3 }$
• B. $\dfrac { [ f ( x ) ] ^ { 3 } } { 3 }$
• C. $[ f ( x ) ] ^ { 2 }$
• D. $\dfrac { 1 } { 2 } [ f ( x ) ] ^ { 2 }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int \cot x {dx}=$
• A. $\ln (\sin^2x) +C$
• B. $(\sin x) +C$
• C. None of these
• D. $\ln (\sin x) +C$

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
$\displaystyle\int{\frac{x}{\sqrt{9+8x-x^2}}dx}$ is equal to
• A. $\displaystyle-\sqrt{9+8x-x^2}+4\cos^{-1}{\left(\frac{x-4}{5}\right)}+C$
• B. $\displaystyle-\sqrt{9+8x-x^2}+4\cos^{-1}{\left(\frac{x-3}{2}\right)}+C$
• C. $\displaystyle\sqrt{9+8x-x^2}+4\sin^{-1}{\left(\frac{x-4}{5}\right)}+C$
• D. $\displaystyle-\sqrt{9+8x-x^2}+4\sin^{-1}{\left(\frac{x-4}{5}\right)}+C$

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 $\displaystyle f(x)>0 \forall x \in R$