Integrals: Exam-Pattern Drill (6)

hard • 2 min read
Tags Integrals

Question

Evaluate

∫xx2+1 dx\int \frac{x}{x^2 + 1}\, dx

Solution — Step by Step

The numerator xx is (up to a constant) the derivative of x2+1x^2 + 1. This is the dead giveaway for uu-substitution.

du=2x dxdu = 2x\, dx, so x dx=12dux\, dx = \tfrac{1}{2}du.

 ∫xx2+1 dx=∫1u⋅12 du=12ln⁡∣u∣+C\int \frac{x}{x^2 + 1}\, dx = \int \frac{1}{u} \cdot \frac{1}{2}\, du = \frac{1}{2}\ln|u| + C =12ln⁡∣x2+1∣+C=12ln⁡(x2+1)+C= \frac{1}{2}\ln|x^2 + 1| + C = \frac{1}{2}\ln(x^2+1) + C

(We drop the absolute value since x2+1>0x^2+1 > 0 always.)

Final answer: 12ln⁡(x2+1)+C\tfrac{1}{2}\ln(x^2+1) + C.

Why This Works

The pattern ∫f′(x)/f(x) dx=ln⁡∣f(x)∣+C\int f'(x)/f(x)\, dx = \ln|f(x)| + C is one of the highest-leverage tricks in CBSE Class 12 integration. Whenever the numerator looks like a derivative of the denominator, the answer is immediately a logarithm.

This single recognition cuts down 30% of CBSE 4-mark integration questions.

Alternative Method

Write xx2+1=12⋅2xx2+1\tfrac{x}{x^2+1} = \tfrac{1}{2} \cdot \tfrac{2x}{x^2+1}. The numerator is now exactly the derivative of the denominator, so the integral is 12ln⁡(x2+1)+C\tfrac{1}{2}\ln(x^2+1) + C in one line.

For ∫1x2+1 dx\int \tfrac{1}{x^2+1}\, dx (no xx in the numerator), the answer is tan⁡−1x+C\tan^{-1} x + C. Don’t confuse the two — the presence or absence of xx in the numerator decides logarithm vs arctangent.

Common Mistake

Writing ∫xx2+1 dx=tan⁡−1(x2+1)+C\int \tfrac{x}{x^2+1}\, dx = \tan^{-1}(x^2+1) + C. This mixes up two different patterns. The arctan formula needs 1x2+1\tfrac{1}{x^2+1} in the integrand (no xx), not xx2+1\tfrac{x}{x^2+1}.

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