Q.Integrate the function $\frac{\sqrt{x^2+1}\,[\log(x^2+1)-2\log x]}{x^4}$

Simplify the logs, then substitute. Since $\log(x^2+1)-2\log x=\log\dfrac{x^2+1}{x^2}=\log\!\left(1+\dfrac1{x^2}\right)$ and (for $x>0$) $\dfrac{\sqrt{x^2+1}}{x^4}=\dfrac{\sqrt{1+1/x^2}}{x^3}$, the integral is

Let $t=1+\dfrac1{x^2}$, so $dt=-\dfrac{2}{x^3}dx$, giving $-\dfrac12\displaystyle\int t^{1/2}\log t\,dt$.

By parts ($u=\log t,\,dv=t^{1/2}dt$): $\int t^{1/2}\log t\,dt=\tfrac23t^{3/2}\log t-\tfrac49t^{3/2}$. Multiplying by $-\tfrac12$ and restoring $t$: …

1. Simplify the logarithm

2. Rewrite the algebraic part

For $x>0$, $\sqrt{x^2+1}=x\sqrt{1+1/x^2}$, so

The integral becomes

3. Substitute

Let $t=1+\dfrac1{x^2}$. Then $dt=-\dfrac{2}{x^3}\,dx$, i.e. $\dfrac{dx}{x^3}=-\dfrac{dt}{2}$, and $\sqrt{1+1/x^2}=\sqrt t$:

4. Integrate by parts

Take $u=\log t$, $dv=t^{1/2}\,dt$, so $du=\dfrac{dt}{t}$, $v=\dfrac23t^{3/2}$: …

Method: Combine logs, factor out the dominant power, substitute, then integrate by parts

Use this for integrands mixing $\sqrt{x^2+1}$, powers of $x$, and $\log$ terms: log-laws and a $t=1+x^{-2}$ substitution reduce it to $\int t^{1/2}\log t\,dt$.

Steps

Step 1: Merge the logarithms.

$\log(x^2+1)-2\log x=\log\dfrac{x^2+1}{x^2}=\log\!\bigl(1+\tfrac1{x^2}\bigr)$.

Step 2: Factor $x$ out of the radical and substitute. …

Common Mistakes

Mistake 1: Not combining the logarithms first.

Why it's wrong: $\log(x^2+1)-2\log x=\log(1+\tfrac1{x^2})$ is the key simplification; skipping it blocks the substitution. Correct approach: use log-laws before anything else.

Mistake 2: Mishandling $\dfrac{\sqrt{x^2+1}}{x^4}$.

Why it's wrong: it must be rewritten as $\dfrac{\sqrt{1+1/x^2}}{x^3}$ so that $t=1+1/x^2$ works. Correct approach: factor $x$ out of the radical. …