Evaluate ∫(0 to 1) x^n · ln(x) dx using integration by parts

Question

Evaluate the definite integral:

\int_0^1 x^n \ln(x) \, dx

where $n > 0$ .

(JEE Advanced 2023, similar pattern)

Solution — Step by Step

Let $u = \ln(x)$ and $dv = x^n \, dx$ .

Then $du = \frac{1}{x} dx$ and $v = \frac{x^{n+1}}{n+1}$ .

Using integration by parts: $\int u \, dv = uv - \int v \, du$ .

\int_0^1 x^n \ln(x) \, dx = \left[\frac{x^{n+1}}{n+1} \cdot \ln(x)\right]_0^1 - \int_0^1 \frac{x^{n+1}}{n+1} \cdot \frac{1}{x} \, dx

= \left[\frac{x^{n+1} \ln(x)}{n+1}\right]_0^1 - \frac{1}{n+1}\int_0^1 x^n \, dx

At $x = 1$ : $\frac{1^{n+1} \cdot \ln(1)}{n+1} = \frac{1 \cdot 0}{n+1} = 0$ .

At $x = 0$ : We need $\lim_{x \to 0^+} \frac{x^{n+1} \ln(x)}{n+1}$ . Since $n > 0$ , $x^{n+1} \to 0$ faster than $\ln(x) \to -\infty$ . By L’Hopital’s rule or direct analysis, this limit is $0$ .

So the boundary term vanishes: $\left[\frac{x^{n+1} \ln(x)}{n+1}\right]_0^1 = 0 - 0 = 0$ .

-\frac{1}{n+1}\int_0^1 x^n \, dx = -\frac{1}{n+1} \cdot \left[\frac{x^{n+1}}{n+1}\right]_0^1 = -\frac{1}{n+1} \cdot \frac{1}{n+1}

= \boxed{-\frac{1}{(n+1)^2}}

Why This Works

The trick is recognising that $\ln(x)$ becomes $-\infty$ as $x \to 0^+$ , but $x^n$ goes to $0$ fast enough to “tame” the logarithm. The product $x^n \ln(x)$ approaches $0$ as $x \to 0^+$ for any $n > 0$ . This is why the boundary term vanishes.

The result $-\frac{1}{(n+1)^2}$ is negative, which makes sense: on $[0, 1]$ , $x^n > 0$ but $\ln(x) < 0$ (since $\ln(x)$ is negative for $0 < x < 1$ ). So the integrand is negative throughout the interval, and the integral must be negative.

This result has a beautiful connection to the Euler-Mascheroni constant and the Gamma function in advanced mathematics.

Alternative Method — Differentiation under the integral sign

Consider $I(n) = \int_0^1 x^n \, dx = \frac{1}{n+1}$ .

Differentiate both sides with respect to $n$ :

\frac{dI}{dn} = \int_0^1 \frac{\partial}{\partial n}(x^n) \, dx = \int_0^1 x^n \ln(x) \, dx

And $\frac{d}{dn}\left(\frac{1}{n+1}\right) = -\frac{1}{(n+1)^2}$ .

This technique — differentiating a known integral with respect to a parameter — is sometimes called Feynman’s trick. It’s extremely powerful for JEE Advanced and often gives results faster than integration by parts. If you see a parameter inside the integrand and a known “base” integral, try this approach.

Common Mistake

Many students panic at the lower limit $x = 0$ because $\ln(0)$ is undefined. They declare the integral “divergent” without checking. But the integral is perfectly convergent — $x^n$ kills the $\ln(x)$ singularity. Always evaluate the limit carefully before concluding that an improper integral diverges.

Question

Solution — Step by Step

Why This Works

Alternative Method — Differentiation under the integral sign

Common Mistake

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