Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals4 Marks
Question
Integrate the rational function $\frac{x}{\left(x^{2}+1\right)(x-1)}$
✓
Answer
Let $\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C}{(x-1)}$ x = (Ax + B)(x - 1) + C (x2 + 1) $\Rightarrow$ x = Ax2 - Ax + Bx - B + Cx2 + C $\Rightarrow$ x = (A+C)x2 - (A-B)x - (B+C) Equating the coefficients of x2, x and constant term, we get, A + C = 0 -A + B = 1 -B + C = 0 On solving these equation, we get, $A=-\frac{1}{2}, B=\frac{1}{2}$ and $C=\frac{1}{2}$ Thus, $\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{\left(-\frac{1}{2} x+\frac{1}{2}\right)}{\left(x^{2}+1\right)}+\frac{\frac{1}{2}}{(x-1)}$ $\Rightarrow~~\int \frac{x}{\left(x^{2}+1\right)(x-1)} d x=-\frac{1}{2} \int \frac{x}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x-1} d x$ = $-\frac{1}{4} \int \frac{2 x}{x^{2}+1} d x+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C$ Now, let us consider, $\int \frac{2 x}{x^{2}+1} d x$ Let (x2 + 1) = t 2xdx = dt Thus, $\int \frac{x}{\left(x^{2}+1\right)(x-1)} d x=-\frac{1}{4} \log \left|\mathrm{x}^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C$ = $\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.