Q.Integrate the following function: $\frac{1}{x (\log x)^m}$, $x>0, m \neq 1$

Concept: U Substitution — the derivative of $\log x$ appears in the denominator, so substitute $u = \log x$.

Let $u = \log x$. Then $du = \frac{1}{x} dx$, and the integral becomes:

Integrate using the power rule (since $m \neq 1$): …

Why substitution works here

When you see a function like $\frac{1}{x (\log x)^m}$, your first instinct should be to look for a composition — something inside something else. Here, the denominator has $x$ multiplied by a power of $\log x$. That $x$ in the denominator is a strong hint: the derivative of $\log x$ is exactly $\frac{1}{x}$. So if we set $u = \log x$, then $du = \frac{1}{x} dx$, and the whole integral collapses into something much simpler.

This is the classic pattern for $u$-substitution: you spot a function and its derivative (up to a constant) appearing together. Here, $\frac{1}{x}$ is the derivative of $\log x$, and $(\log x)^m$ is the function itself raised to a power. That’s a perfect match.

Step-by-step solution

1. Set up the substitution. Let $u = \log x$. Then differentiate:

2. Rewrite the integral in terms of $u$. The original integral is $\int \frac{1}{x (\log x)^m} \, dx = \int \frac{1}{(\log x)^m} \cdot \frac{1}{x} dx.$ …

Method: Substitute $u=\log x$ for a power of a logarithm over $x$

Use this when $(\log x)$ is raised to a power and divided by $x$, e.g. $\dfrac{1}{x(\log x)^m}$: the $\dfrac1x$ is precisely $d(\log x)$.

Steps

Step 1: Substitute the logarithm.

Let $u=\log x$, so $du=\dfrac1x\,dx$. The integral becomes $\displaystyle\int u^{-m}\,du$.

Step 2: Apply the power rule (note $m\neq1$).

$\int u^{-m}\,du=\frac{u^{-m+1}}{-m+1}+C=\frac{u^{1-m}}{1-m}+C.$ …

Common Mistakes

Mistake 1: Using the log rule instead of the power rule.

Why it's wrong: after $u=\log x$ the integral is $\int u^{-m}\,du$; since $m\neq1$, the exponent $-m\neq-1$, so it is a power, not a logarithm. Correct approach: apply $\dfrac{u^{1-m}}{1-m}$, reserving $\log|u|$ only for $m=1$.

Mistake 2: Sign/arithmetic slip in the exponent $1-m$. …