Definite Integration
Definite + Partial Fractions
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^1\frac{x}{(x^2+1)^2}\,dx\) [JEE Main 2020]</p>
<li>\(\dfrac{1}{4}\)</li>
<li>\(\dfrac{1}{2}\)</li>
<li>\(\dfrac{\pi}{4}\)</li>
<li>\(1\)</li>
Step-by-Step Solution
Key Concept: Let t = x^2+1, dt = 2x dx. \int = (1/2)\int_1^2 t⁻^2 dt = (1/2)[-1/t]_1^2 = (1/2)(-1/2+1) = 1/4.
<div class='solution'>
<p>Let $t=x^2+1\Rightarrow dt=2x\,dx$. Limits: $x=0\to t=1$; $x=1\to t=2$.</p>
<p>$$I=\frac{1}{2}\int_1^2 t^{-2}\,dt=\frac{1}{2}\left[-\frac{1}{t}\right]_1^2=\frac{1}{2}\left(-\frac{1}{2}+1\right)=\boxed{\frac{1}{4}}$$</p>
</div>
Correct Answer: A