The value of $\int \frac{\mathrm{d} x}{7+6 x-x^2}$ is equal to

The value of $\int \frac{\mathrm{d} x}{7+6 x-x^2}$ is equal to
  1. $\frac{1}{4} \log \left(\frac{1+x}{7-x}\right)+\mathrm{c}$, (where c is a constant of integration)
  2. $\frac{1}{8} \log \left(\frac{7-x}{1+x}\right)+\mathrm{c},($ where c is a constant of integration)
  3. $\frac{1}{4} \log \left(\frac{7-x}{1+x}\right)+\mathrm{c}$, (where c is a constant of integration)
  4. $\frac{1}{8} \log \left(\frac{1+x}{7-x}\right)+\mathrm{c}$, (where c is a constant of integration)

Solution

$\begin{aligned} \int \frac{1}{7+6 x-x^2} \mathrm{~d} x & =\int \frac{1}{7+6 x-x^2-9+9} \mathrm{~d} x \\ & =\int \frac{1}{16-\left(x^2-6 x+9\right)} \mathrm{d} x \\ & =\int \frac{1}{4^2-(x-3)^2} \mathrm{~d} x \\ & =\frac{1}{2(4)} \log \left|\frac{4+(x-3)}{4-(x-3)}\right|+\mathrm{c} \\ & =\frac{1}{8} \log \left|\frac{1+x}{7-x}\right|+\mathrm{c}\end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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