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Exercise: Integration by Substitution · Q12

Q.Evaluate ∫xx2+1 dx\displaystyle\int \frac{x}{x^2+1}\,dx.

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Concept understanding — Integration by Substitution

The substitution (change-of-variable) method mirrors the chain rule of differentiation. If u=g(x)u=g(x) is a differentiable function, then

∫f(g(x)) g′(x) dx=∫f(u) du,\int f\big(g(x)\big)\,g'(x)\,dx = \int f(u)\,du,

because du=g′(x) dxdu=g'(x)\,dx. Choosing uu so that its derivative already appears (up to a constant) in the integrand converts a hard integral into a standard one; after integrating in uu, substitute back u=g(x)u=g(x).

Two especially useful consequences (with u=f(x)u=f(x)):

∫f′(x)f(x) dx=log⁡∣f(x)∣+c,∫f′(x) [f(x)]n dx=[f(x)]n+1n+1+c (n≠−1).\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)|+c,\qquad \int f'(x)\,[f(x)]^{n}\,dx = \frac{[f(x)]^{n+1}}{n+1}+c\ (n\ne -1).

Standard log-form results that follow are ∫tan⁡x dx=log⁡∣sec⁡x∣+c\int \tan x\,dx=\log|\sec x|+c, ∫cot⁡x dx=log⁡∣sin⁡x∣+c\int \cot x\,dx=\log|\sin x|+c, ∫cosec⁡x dx=log⁡∣cosec⁡x−cot⁡x∣+c\int \operatorname{cosec} x\,dx=\log|\operatorname{cosec} x-\cot x|+c, and ∫sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣+c\int \sec x\,dx=\log|\sec x+\tan x|+c.

For a trigonometric substitution (e.g. x=atan⁡θx=a\tan\theta), draw a right triangle to read back the other trig ratios when reversing the substitution.

Tip

The whole method rests on picking a uu whose differential g′(x) dxg'(x)\,dx is present in the integrand. If it isn't (even up to a constant multiple), substitution won't simplify things — try a different method.

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