# Math Help - Integral

1. ## Integral

find out the value of the integration by using partial fraction knowledge, $\int_{0}^{k}\frac{2}{(x+1)^2(x^2+1)}d(x)$ where [k>0]then $\lim_{k\rightarrow\propto}\int_{0}^{k}\frac{2}{(x+ 1)^2(x^2+1)}d(x)$,,,,I go through this way,,, first of all i find the value of the integral, it comes like this, $\frac{1}{2}ln\frac{(k+1)^2}{k^2+1}+\frac{k}{k+1}$then can i change intho this and find out the limit for that,, $\frac{1}{2}ln\frac{(\frac{1}{k}+1)^2}{\frac{1}{k^2 }+1}+\frac{1}{\frac{1}{k}+1}$ Is It Correct or Not???

2. ## Re: Integral

Hello, srirahulan!

Evaulate the integral with partial fractions: . $\int_{0}^{k}\frac{2\,dx}{(x+1)^2(x^2+1)}$ where $k>0$
then find: / $\lim_{k\to\infty}\int_{0}^{k}\frac{2\,dx}{(x+1)^2( x^2+1)}$

We have: . $\frac{2}{(x+1)^2(x^2+1)} \;=\;\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{Cx}{x^2+1} + \frac{D}{x^2+1}$

. . $2 \;=\;A(x+1)(x^2+1) + B(x^2+1) + Cx(x+1)^2 D(x+1)^2$

$\text{Let }x = \text{-}1\!:\;2 \;=\;A(0) + B(2) + C(0) + D(0) \quad\Rightarrow\quad \boxed{B = 1}\;\;[1]$

$\text{Let }x = 0\!:\;2 \;=\;A(1) + B(1) + C(0) + D(1)\;\;[2]$

$\text{Let }x = 1\!:\;2 \;=\;A(4) + B(2) + C(4) + D(4)\;\;[3]$

$\text{Let }x = 2\!:\;2 \;=\;A(15) + B(5) + C(18) + D(9)\;\;[4]$

$\begin{array}{cccccccc}\text{[2] becomes:} & A + 1 + D &=&2 & \Rightarrow & A+D &=& 1 \\ \text{[3] becomes:} & 4A+ 2 + 4C + 4D &=& 2 & \Rightarrow & A+C+D &=& 0 \\ \text{[4] becomes:} & 15A + 5 + 18C + 9D &=& 2 & \Rightarrow & 5A + 6B + 3B &=& \text{-}1 \end{array}$

$\text{Solve the system: }\:\begin{Bmatrix}A&=&1 \\ B&=&1 \\ C&=&\text{-}1 \\ D&=&0 \end{Bmatrix}$

$\text{Hence: }\:\int^k_0\frac{2\,dx}{(x+1)^2(x^2+1)} \;=\;\int^k_0\left[\frac{1}{x+1} + \frac{1}{(x+1)^2} - \frac{x}{x^2+1}\right]\,dx$

. . . $=\;\ln(x+1) - \frac{1}{x+1} - \tfrac{1}{2}\ln|x^2+1|\,\bigg]^k_0$

. . . $=\;\tfrac{1}{2}\left[2\ln(x+1) - \frac{2}{1+x} - \ln(x^2+1)\right]^k_0$

. . . $=\;\tfrac{1}{2}\left[\ln\frac{(x+1)^2}{x^2+1} = \frac{2}{x+1}\right]^k_0$

Can you continue?

4. ## Re: Integral

Originally Posted by Soroban
Hello, srirahulan!

We have: .$$\frac{2}{(x+1)^2(x^2+1)} \;=\;\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{Cx}{x^2+1} + \frac{D}{x^2+1}$$

. . $$2 \;=\;A(x+1)(x^2+1) + B(x^2+1) + Cx(x+1)^2 D(x+1)^2$$

$$\text{Let }x = \text{-}1\!:\;2 \;=\;A(0) + B(2) + C(0) + D(0) \quad\Rightarrow\quad \boxed{B = 1}\;\;[1]$$

$$\text{Let }x = 0\!:\;2 \;=\;A(1) + B(1) + C(0) + D(1)\;\;[2]$$

$$\text{Let }x = 1\!:\;2 \;=\;A(4) + B(2) + C(4) + D(4)\;\;[3]$$

$$\text{Let }x = 2\!:\;2 \;=\;A(15) + B(5) + C(18) + D(9)\;\;[4]$$

$$\begin{array}{cccccccc}\text{[2] becomes:} & A + 1 + D &=&2 & \Rightarrow & A+D &=& 1 \\ \text{[3] becomes:} & 4A+ 2 + 4C + 4D &=& 2 & \Rightarrow & A+C+D &=& 0 \\ \text{[4] becomes:} & 15A + 5 + 18C + 9D &=& 2 & \Rightarrow & 5A + 6B + 3B &=& \text{-}1 \end{array}$$

$$\text{Solve the system: }\:\begin{Bmatrix}A&=&1 \\ B&=&1 \\ C&=&\text{-}1 \\ D&=&0 \end{Bmatrix}$$

$$\text{Hence: }\:\int^k_0\frac{2\,dx}{(x+1)^2(x^2+1)} \;=\;\int^k_0\left[\frac{1}{x+1} + \frac{1}{(x+1)^2} - \frac{x}{x^2+1}\right]\,dx$$

. . . $$=\;\ln(x+1) - \frac{1}{x+1} - \tfrac{1}{2}\ln|x^2+1|\,\bigg]^k_0$$

. . . $$=\;\tfrac{1}{2}\left[2\ln(x+1) - \frac{2}{1+x} - \ln(x^2+1)\right]^k_0$$

. . . $$=\;\tfrac{1}{2}\left[\ln\frac{(x+1)^2}{x^2+1} = \frac{2}{x+1}\right]^k_0$$

Can you continue?
redone for new latex

5. ## Re: Integral

Originally Posted by srirahulan
find out the value of the integration by using partial fraction knowledge,$$\int_{0}^{k}\frac{2}{(x+1)^2(x^2+1)}d( x)$$ where [k>0]then $$\lim_{k\rightarrow\propto}\int_{0}^{k}\frac{2}{( x+1)^2(x^2+1)}d(x)$$,,,,I go through this way,,, first of all i find the value of the integral, it comes like this, $$\frac{1}{2}ln\frac{(k+1)^2}{k^2+1}+\frac{k}{k+1}$$then can i change intho this and find out the limit for that,,$$\frac{1}{2}ln\frac{(\frac{1}{k}+1)^2}{ \frac{1}{k^2}+1}+\frac{1}{\frac{1}{k}+1}$$ Is It Correct or Not???
redone for new LaTex

6. ## Re: Integral

OP, yes, you can change it just as you suggested.

7. ## Re: Integral

okay i got that, Thanks for everybody>>>>>>