$\int \frac{x^3-7 x+6}{x^2+3 x} \mathrm{~d} x=$
$\int \frac{x^3-7 x+6}{x^2+3 x} \mathrm{~d} x=$
- $\frac{x^2}{2}+3 x-\log |x|+\mathrm{c}$, where c is a constant of integration.
- $\frac{x^2}{2}+3 x+2 \log |x|+\mathrm{c}$, where c is a constant of integration.
- $\frac{x^2}{2}-3 x+2 \log |x|+\mathrm{c}$, where c is a constant of integration.
- $\frac{x^2}{2}-3 x-\log |x|+\mathrm{c}$, where c is a constant of integration.
Solution
$\begin{aligned} \text {Let I } & =\int \frac{x^3-7 x+6}{x^2+3 x} \\ & =\int\left(x-3+\frac{2 x+6}{x^2+3 x}\right) \mathrm{d} x \\ & =\int\left(x-3+\frac{2(x+3)}{x(x+3)}\right) \mathrm{d} x \\ & =\int\left(x-3+\frac{2}{x}\right) \mathrm{d} x \\ & =\frac{x^2}{2}-3 x+2 \log |x|+\mathrm{c}\end{aligned}$
Asked in: MHT CET 2024 (16 May Shift 2)
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