Search any question & find its solution
Question:
Answered & Verified by Expert
The intearrati the differential equation $\left(1+x^{2}\right) d t=\left(\tan ^{-1} x-t\right) d x$
Options:
Solution:
1991 Upvotes
Verified Answer
The correct answer is:
$e^{\tan ^{-1} x}$
(D)
$\begin{aligned}
\frac{\mathrm{dt}}{\mathrm{dx}} &=\frac{\tan ^{-1}-\mathrm{t}}{1+\mathrm{x}^{2}} \\
\therefore \frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{t}}{1+\mathrm{x}^{2}} &=\frac{\tan ^{-1} \mathrm{x}}{1+\mathrm{x}^{2}} \\
\text { I.F. }=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}} &=\mathrm{e}^{\tan ^{-1} \mathrm{x}}
\end{aligned}$
$\begin{aligned}
\frac{\mathrm{dt}}{\mathrm{dx}} &=\frac{\tan ^{-1}-\mathrm{t}}{1+\mathrm{x}^{2}} \\
\therefore \frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{t}}{1+\mathrm{x}^{2}} &=\frac{\tan ^{-1} \mathrm{x}}{1+\mathrm{x}^{2}} \\
\text { I.F. }=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}} &=\mathrm{e}^{\tan ^{-1} \mathrm{x}}
\end{aligned}$
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.