Search any question & find its solution
Question:
Answered & Verified by Expert
$\int \frac{1}{7-6 x-x^2} d x=$
Options:
Solution:
2647 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{8} \log \left(\frac{7+x}{1-x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
Let
$\begin{aligned}
\mathbf{I} & =\int \frac{1}{7-6 x-x^2} \mathrm{~d} x \\
& =\int \frac{1}{7-6 x-x^2-9+9} \mathrm{~d} x \\
& =\int \frac{1}{16-\left(x^2+6 x+9\right)} \mathrm{d} x \\
& =\int \frac{1}{(4)^2-(x+3)^2} \mathrm{~d} x \\
& =\frac{1}{8} \log \left|\frac{4+x+3}{4-(x+3)}\right|+\mathrm{c} \\
& =\frac{1}{8} \log \left|\frac{7+x}{1-x}\right|+\mathrm{c}
\end{aligned}$
$\begin{aligned}
\mathbf{I} & =\int \frac{1}{7-6 x-x^2} \mathrm{~d} x \\
& =\int \frac{1}{7-6 x-x^2-9+9} \mathrm{~d} x \\
& =\int \frac{1}{16-\left(x^2+6 x+9\right)} \mathrm{d} x \\
& =\int \frac{1}{(4)^2-(x+3)^2} \mathrm{~d} x \\
& =\frac{1}{8} \log \left|\frac{4+x+3}{4-(x+3)}\right|+\mathrm{c} \\
& =\frac{1}{8} \log \left|\frac{7+x}{1-x}\right|+\mathrm{c}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.