Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Find: $\int \frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)} d x$
✓
Answer
Let, $I=\int \frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)} d x$
By partial fractions
$\frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+1}$
$x^2 + x + 1 = A(x^2+ 1) + Bx + C(x + 2)$
$=Ax^2+ A + Bx^2+ 2Bx + Cx + 2Cx^2+ x + 1$
$= x^2(A + B) + x(2B + C) + A + 2C$
On comparing coefficients of equation $(i),$ we get
$1 = A + B$ and $1 = 2B + C$
On solving above three equations, we get
$A=\frac{3}{5}$, $ B=\frac{2}{5} $ and $C=\frac{1}{5}$
Hence, $\frac{x^{2}+x+1}{(x+2)\left(x^{2}+1\right)}=\frac{3}{5(x+2)}+\frac{\frac{2}{5} x+\frac{1}{5}}{x^{2}+1}$
Therefore, $\int \frac{\left(x^{2}+x+1\right) d x}{(x+2)\left(x^{2}+1\right)}=\frac{3}{5} \int \frac{1}{x+2} d x+\frac{2}{5} \int \frac{x}{x^{2}+1} d x +\frac{1}{5} \int \frac{1}{\left(x^{2}+1\right)} d x$
= $\frac{3}{5} \log |x+2|+\frac{1}{5} \int \frac{2 x}{x^{2}+1}+\frac{1}{5} \int \frac{d x}{x^{2}+1}$
= $\frac{3}{5} \log |x+2|+\frac{1}{5} \log \left|x^{2}+1\right|+\frac{1}{5} \tan ^{-1} x+c$
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