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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration4 Marks
Question
Evaluate : $\int \frac{3 x^2+4 x-5}{\left(x^2-1\right)(x+2)} d x$
✓

Answer

$\mathrm{I}=\int \frac{3 x^2+4 x-5}{(x-1)(x+1)(x+2)} \cdot d x$
$
\begin{aligned}
\text { Consider, } \quad \frac{3 x^2+4 x-5}{(x-1)(x+1)(x+2)} & =\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(x+2)} \\
& =\frac{A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)}{(x-1)(x+1)(x+2)} \\
\therefore \quad 3 x^2+4 x-5=A(x+1)(x+2) & +B(x-1)(x+2)+C(x-1)(x+1) \\
\text { at } x=1, \quad 3(1)^2+4(1)-5 & =A(2)(3)+B(0)+C(0) \\
2=6 A \quad \Rightarrow A & =\frac{1}{3} \\
\text { at } x=-1, \quad 3(-1)^2+4(-1)-5 & =A(0)+B(-2)(1)+C(0) \\
-6=-2 B \Rightarrow B & =3
\end{aligned}
$
at $x=1$,
$
3(1)^2+4(1)-5=A(2)(3)+B(0)+C(0)
$
at $x=-1$,
$
3(-1)^2+4(-1)-5=A(0)+B(-2)(1)+C(0)
$
at $x=-2, \quad 3(-2)^2+4(-2)-5=A(0)+B(0)+C(-3)(-1)$
$
-1=3 C \quad \Rightarrow \quad C=-\frac{1}{3}
$
Thus, $\quad \frac{3 x^2+4 x-5}{(x-1)(x+1)(x+2)}=\frac{\left(\frac{1}{3}\right)}{(x-1)}+\frac{3}{(x+1)}+\frac{\left(-\frac{1}{3}\right)}{(x+2)}$
$
\begin{aligned}
\therefore \mathrm{I} & =\int\left[\frac{\left(\frac{1}{3}\right)}{(x-1)}+\frac{3}{(x+1)}+\frac{\left(-\frac{1}{3}\right)}{(x+2)}\right] \cdot d x & =\frac{1}{3} \log (x-1)+3 \log (x+1)-\frac{1}{3} \log (x+2)+c \\
& =\frac{1}{3} \log \left[\frac{(x-1)(x+1)^9}{(x+2)}\right]+c & \therefore \int \frac{3 x^2+4 x-5}{\left(x^2-1\right)(x+2)} \cdot d x=\frac{1}{3} \log \left[\frac{(x-1)(x+1)^9}{(x+2)}\right]+c
\end{aligned}
$

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