Search any question & find its solution
Question:
Answered & Verified by Expert
The integral of $\frac{x^2-x}{x^3-x^2+x-1}$ w.r.t. $x$ is
Options:
Solution:
2051 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2} \log \left(x^2+1+c\right)$
$\frac{1}{2} \log \left(x^2+1+c\right)$
Let $I=\int \frac{x^2-x}{x^3-x^2+x-1} d x$
$$
=\int \frac{x(x-1)}{x^2(x-1)+(x-1)} d x=\int \frac{x d x}{x^2+1}
$$
$$
\left.=\frac{1}{2} \int \frac{2 x d x}{\left(x^2+1\right.}\right)
$$
Let $x^2+1=t \Rightarrow 2 x d x=d t$
$$
\begin{aligned}
& \therefore I=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log t+c \\
& =\frac{1}{2} \log \left(x^2+1\right)+c
\end{aligned}
$$
where ' $c$ ' is the constant of integration.
$$
=\int \frac{x(x-1)}{x^2(x-1)+(x-1)} d x=\int \frac{x d x}{x^2+1}
$$
$$
\left.=\frac{1}{2} \int \frac{2 x d x}{\left(x^2+1\right.}\right)
$$
Let $x^2+1=t \Rightarrow 2 x d x=d t$
$$
\begin{aligned}
& \therefore I=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log t+c \\
& =\frac{1}{2} \log \left(x^2+1\right)+c
\end{aligned}
$$
where ' $c$ ' is the constant of integration.
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.