Evaluate the following integrals:


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


$=\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$

by partial fraction,


So we get these three equations,

$2 \mathrm{~A}+2 \mathrm{~B}+\mathrm{C}=1$

$3 \mathrm{~A}+\mathrm{B}+2 \mathrm{C}=1$


So the values are $A=-2 ; C=3 ; B=1$

$\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x=\int\left(-\frac{2 d x}{x+1}\right)+\int \frac{d x}{(x+1)^{2}}+\int \frac{3 d x}{x+2}$

$=-2 \log (x+1)+3 \log (x+2)-\frac{1}{x+1}+c$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now