# Thread: integration of (2x+3)/(x+1)^2 for 0 to 1 by partial fractions

1. ## integration of (2x+3)/(x+1)^2 for 0 to 1 by partial fractions

Question is to evaluate by partial fractions
$\displaystyle \int_0^1 {\frac{{2x + 3}}{{(x + 1)^2 }}} dx$

my attempt

$\displaystyle \begin{array}{l} \frac{{2x + 3}}{{(x + 1)(x + 1)}} = \frac{A}{{x + 1}} + \frac{B}{{(x + 1)^2 }} \\ \frac{{2x + 3}}{{(x + 1)^2 }} = A(x + 1) + B \\ 2x + 3 = A(x) + (A + B) \\ \end{array}$

giving coefficients as

$\displaystyle \begin{array}{l} A = 2 \\ A + B = 3 \\ so \\ B = 1 \\ \end{array}$

and so the original integral is equal to

$\displaystyle \int_0^1 {2\frac{1}{{x + 1}}dx} + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx}$

$\displaystyle = 2\left[ {\ln |x + 1|} \right]_0^1 + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx}$

I think i'm tripping up here on the 2nd integral
I had
$\displaystyle \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = \left[ {\frac{1}{2}\ln (x + 1)} \right]_0^1$

But I don't think this is right. Can someone show me how to integrate the 2nd integral? Thanks.

2. $\displaystyle \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = \left[ {\frac{-1}{x+1}} \right]_0^1$

Try substituting x+1 = u
dx= du
So basically you need to integrate 1/u^2
-after integrating substitute u by x+1
-then put the limits

3. thanks

$\displaystyle \begin{array}{l} \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = u^{ - 2} dx \\ u = x + 1 \\ du = 1dx \\ \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = u^{ - 2} du \\ \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = \left[ {\frac{{u^{ - 1} }}{{ - 1}}} \right]_0^1 = \left[ {\frac{{ - 1}}{{x + 1}}} \right]_0^1 \\ \end{array}$

and so

$\displaystyle \begin{array}{l} \int_0^1 {2\frac{1}{{x + 1}}dx} + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} \\ = 2\left[ {\ln |x + 1|} \right]_0^1 + \left[ {\frac{{ - 1}}{{x + 1}}} \right]_0^1 \\ = 2\ln 2 + \frac{1}{2} \\ \end{array}$

4. Originally Posted by Craka
thanks

$\displaystyle \begin{array}{l} \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = u^{ - 2} dx \\ u = x + 1 \\ du = 1dx \\ \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = u^{ - 2} du \\ \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} = \left[ {\frac{{u^{ - 1} }}{{ - 1}}} \right]_0^1 = \left[ {\frac{{ - 1}}{{x + 1}}} \right]_0^1 \\ \end{array}$

and so

$\displaystyle \begin{array}{l} \int_0^1 {2\frac{1}{{x + 1}}dx} + \int_0^1 {1\frac{1}{{(x + 1)^2 }}dx} \\ = 2\left[ {\ln |x + 1|} \right]_0^1 + \left[ {\frac{{ - 1}}{{x + 1}}} \right]_0^1 \\ = 2\ln 2 + \frac{1}{2} \\ \end{array}$