Mathematics

# Evaluate: $\displaystyle \int{\dfrac{(2x+3)dx}{x^2+3x+6}}$

##### SOLUTION
$\displaystyle \int{\dfrac{(2x+3)dx}{x^2+3x+6}}\\t=x^2+3x+6\implies dt=2x+3dx\\\displaystyle \int \dfrac{dt}{t}\\\log t+c\\\log(x^2+3x+6)+c$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Medium
Solve : $\int \left(\sqrt{x} - \dfrac{1}{\sqrt{x}} \right)^2$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate: $\displaystyle \int{\dfrac{(2x+3)dx}{x^2+3x+6}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\int _{ 0 }^{ \infty }{ \cfrac { \log { x } }{ 1+{ x }^{ 2 } } } dx$, equals
• A. $\cfrac{\pi}{2}$ $\log{2}$
• B. $-\cfrac{\pi}{2}$ $\log{2}$
• C. $\cfrac{\pi}{4}$ $\log{2}$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\int \dfrac {x^2+1}{x^4+1} dx=$ ?
• A. $\frac {1}{\sqrt{2}}tan^{-1}\left ( x-\frac {1}{x} \right )+C$
• B. $\frac {1}{\sqrt{2}}cot^{-1}\left ( x-\frac {1}{x} \right )+C$
• C. none of these
• D. $\frac {1}{\sqrt{2}}tan^{-1}\left \{ \frac {1}{\sqrt{2}} \left ( x-\frac {1}{x} \right ) \right \} +C$

$\int\limits_{ - 1}^1 {{e^x}dx} =$