Mathematics

# Integrate :- $\displaystyle \int_{}^{} {\frac{{dx}}{{x\left( {x + 1} \right)}}}$

##### SOLUTION
$=\int \frac{dx}{x(x+1)}$
= Breaking it into partial fractions.
$\int \frac{dx}{x(x+1)}= A\int \frac{dx}{x}+ B\int \frac{dx}{x+1}$
$1=A(x+1)+Bx$
Put x=0,    put x=-1
A=1           B=-1
$log |x|-log |x+1|+c$
$= log |\frac{x}{x+1}|+c$

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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
If a continuous function $f$ satisfies $\displaystyle \int_{0}^{x^{2}}f\left ( t \right )\: dt= x^{2}\left ( 1+x \right )$ then $f\left ( 4 \right )$ is equal to
• A. $7$
• B. $5$
• C. $6$
• D. $4$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the integral
$\displaystyle \int _{ 4 }^{ 5 }{ \left( { x }^{ 2 }-12x+8 \right) dx }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
A cubic function $\displaystyle f(x)$ vanishes at $\displaystyle x=-2$ and has a relative minima/maxima at $\displaystyle x=1$ and $\displaystyle x=1/3$ if $\displaystyle \int_{-1}^{1}f\left ( x \right )dx= \frac{14}{3}$ then $\displaystyle f(x)$ equals
• A. $\displaystyle \left ( x+2 \right )\left ( x^{2}+x-1 \right )$
• B. $\displaystyle \left ( x+3 \right )\left ( x+2 \right )\left ( x-1 \right )$
• C. None of these
• D. $\displaystyle \left ( x+2 \right )\left ( x^{2}-x+1 \right )$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $I_{1} = \int_{0}^{1} 2^{x^{3}} dx, I_{2} = \int_{0}^{1}2^{x^{2}}dx, I_{3} = \int_{1}^{2}2^{x^{2}}dx$ and $I_{4} = \int_{1}^{2}2^{x^{3}}dx$, then
• A. $I_{2} > I_{1}$
• B. $I_{3} > I_{4}$
• C. $I_{1} > I_{3}$
• D. $I_{1} > I_{2}$

Let $\displaystyle 2I_{1}+I_{2}=\int \frac {e^{x}}{e^{2x}+e^{-2x}}dx$  and  $\displaystyle I_{1}+2I_{2}=\int \frac {e^{-x}}{e^{2x}+e^{-2x}}dx$