Mathematics

# $\int \frac{(x^{2}+2)}{x+1} dx$

##### SOLUTION
$\int \frac{x^{2}+2}{x+1}.dx$ $= \int \left ( x-1 + \frac{3}{x+1} \right ).dx$
$= \int \left ( x-1 \right ).dx + 3 \int \frac{1}{x+1}.dx = \frac{x^{2}}{2} - x + 3\, log \,|x+1|$
$= \frac{x^{2}}{2} - x + 3\, log\, |x+1| + c$

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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
Evaluate: $\displaystyle \int (1+x -x^{-1})e^{x+x^{-1}}dx$
• A. $(x +1)e^{x+x^{-1}}+c$
• B. $(x -1)e^{x+x^{-1}}+c$
• C. $-xe^{x+x^{-1}}+c$
• D. $xe^{x+x^{-1}} + c$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $\displaystyle I=\int\frac{dx}{\sqrt{\left ( x-\alpha \right )\left ( \beta -x \right )}},\left ( \beta < \alpha \right )$ then value of I is

• A. $\displaystyle \sin^{-1}\left ( \frac{x+\alpha+\beta }{\beta -\alpha } \right )$
• B. $\displaystyle \sin^{-1}\left ( \frac{2x+\beta-\alpha }{\beta +\alpha } \right )+C$
• C. none of these
• D. $\displaystyle \sin^{-1}\left ( \frac{2x-\alpha -\beta }{\beta -\alpha } \right )+C$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The following integral $\displaystyle \int_{\pi/4}^{\pi/2} (2 cosec x)^{17}dx$ is equal to
• A. $\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u + e^{-u})^{17} du$
• B. $\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u - e^{-u})^{17} du$
• C. $\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u - e^{-u})^{16} du$
• D. $\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate $\int \int_{R} e^{-(x^{2} + y^{2})}dx dy$, where $R$ is the region bounded by the circle $x^{2} + y^{2} = a^{2}$.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Medium
The average value of a function f(x) over the interval, [a,b] is the number $\displaystyle \mu =\frac{1}{b-a}\int_{a}^{b}f\left ( x \right )dx$
The square root $\displaystyle \left \{ \frac{1}{b-a}\int_{a}^{b}\left [ f\left ( x \right ) \right ]^{2}dx \right \}^{1/2}$ is called the root mean square of f on [a, b]. The average value of $\displaystyle \mu$ is attained id f is continuous on [a, b].

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020