Mathematics

# $\int \frac{x+2}{x^{2}+2x+5}$ is equal to

##### SOLUTION
$\displaystyle\int \dfrac{x+2}{x^2+2x+5}$
$x+2=\lambda\dfrac{d}{dx}(x^2+2x+5)+\mu$
$x+2=2\lambda x+2\lambda +\mu$
$2\lambda =1$, $2\lambda +\mu =2$
$\lambda =\dfrac{1}{2}$, $\mu =1$
$\dfrac{x+2}{x^2+2x+5}=\dfrac{1}{2}\dfrac{(2x+2)}{x^2+2x+5}+\dfrac{1}{x^2+2x+5}$
$\displaystyle\int \dfrac{x+2}{x^2+2x+5}=\dfrac{1}{2}log|x^2+2x+5|+\displaystyle\int \dfrac{1}{x^2+2x+5}$
$=\dfrac{1}{2}log|x^2+2x+5|+\displaystyle\int \dfrac{1}{(x+1)^2+(2)^2}$
$=\dfrac{1}{2} log |x^2+2x+5|+\dfrac{1}{2}\tan^{-1}\left(\dfrac{x+1}{2}\right)$.

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 Hard
$\displaystyle \int _{ 0 }^{ 1 }{ x{ \left( \tan ^{ -1 }{ x } \right) }^{ 2 } } dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\underset {n\rightarrow \infty}{\lim}\dfrac {1}{n}\displaystyle \sum_{r = 1}^{r = 2n} \dfrac {r}{\sqrt {n^{2} + r^{2}}}$ equals.
• A. $\sqrt {5} + 1$
• B. $1 = \sqrt {5}$
• C. None of these
• D. $\sqrt {5} - 1$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
lf $\displaystyle \int f(x)dx=g(x)$ then $\displaystyle \int x^{3}f(x^{2})dx=$
• A. $\dfrac{1}{2}\displaystyle \{x^{2}g(x^{2})-\int g(x^{2})dx\}$
• B. $\dfrac{1}{2}\displaystyle \{x^{2}g(x^{2})+\int g(x^{2})dx^{2}\}$
• C. $\dfrac{1}{2}\displaystyle \{x^{2}g(x^{2})+\int g(x^{2})dx\}$
• D. $\dfrac{1}{2}\displaystyle \{x^{2}g(x^{2})-\int g(x^{2})dx^{2}\}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate: $\displaystyle \int { \cfrac { \sin { \left( \log { x } \right) } }{ x } } dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
In calculating a number of integrals we had to use the method of integration by parts several times in succession.
The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts
$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$
where  $\displaystyle v_{1}\left ( x \right )=\int v\left ( x \right )dx,v_{2}\left ( x \right )=\int v_{1}\left ( x \right )dx ..., v_{n}\left ( x \right )= \int v_{n-1}\left ( x \right )dx$
Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when  calculating $\displaystyle \int P_{n}\left ( x \right )Q\left ( x \right )dx,$ where $\displaystyle P_{n}\left ( x \right )$ is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n+1 times.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020