Q.Evaluate: $\int \dfrac{\sqrt{1+x^2}}{x^4}\,dx$

Key idea: rewrite the root so the integrand becomes a power times its own derivative.

For $x>0$, $\dfrac{\sqrt{1+x^2}}{x^4}=\dfrac{1}{x^3}\cdot\dfrac{\sqrt{1+x^2}}{x}=\dfrac{1}{x^3}\sqrt{\dfrac{1+x^2}{x^2}}=x^{-3}\sqrt{1+x^{-2}}.$

Let $u=1+x^{-2}$, so $du=-2x^{-3}\,dx$, i.e. $x^{-3}\,dx=-\tfrac12\,du$:

$\int x^{-3}\sqrt{1+x^{-2}}\,dx=-\frac12\int u^{1/2}\,du=-\frac12\cdot\frac23 u^{3/2}=-\frac13 u^{3/2}.$ …

Intuition. The denominator $x^4$ is large, so we try to reshape the integrand into "a power of something times the derivative of that something." Splitting off one $x$ from $x^4$ and tucking it under the root does exactly that.

1. Reshape the integrand

For $x>0$,

2. Choose the substitution

Let $u=1+x^{-2}$. Then

The integrand is precisely $\sqrt{u}\cdot x^{-3}\,dx$, so every $x$ is absorbed:

3. Integrate and return to $x$ …

Method: Substituting to expose a power-rule integral

Use this for integrands like $\dfrac{\sqrt{1+x^2}}{x^4}$, where pulling $x$ out of the root (or substituting $t=1+x^{-2}$) reveals a simple $u^{n}\,du$ form.

Steps

Step 1: Factor $x^2$ out of the root to create a negative power.

so $\dfrac{\sqrt{1+x^2}}{x^4}=\dfrac{\sqrt{1+x^{-2}}}{x^3}=x^{-3}\sqrt{1+x^{-2}}$.

Step 2: Substitute $t=1+x^{-2}$.

Step 3: Integrate the power form. …

Common Mistakes

Mistake 1: Trying $u=1+x^2$ directly.

Why it's wrong: $du=2x\,dx$ needs an $x$ in the numerator, but the $x$'s are in the denominator here. Correct approach: factor $x^2$ from the root so $t=1+x^{-2}$ matches the available $x^{-3}\,dx$.

Mistake 2: Sign/constant error from $dt$.

Why it's wrong: $dt=-2x^{-3}\,dx$, so $x^{-3}\,dx=-\tfrac12\,dt$; missing the $-\tfrac12$ gives the wrong coefficient/sign. Correct approach: carry the full $dt$ relationship.

Mistake 3: Not converting $\sqrt{1+x^{-2}}$ back cleanly. …