Q.∫ 𝒅𝒙 𝒙𝟑(𝟏+𝒙𝟒) 𝟏 𝟐 equals
(A) − 1 2𝑥2 √1 + 𝑥4 + 𝑐
(B) 1 2𝑥 √1 + 𝑥4 + 𝑐
(C) − 1 4𝑥 √1 + 𝑥4 + 𝑐
(D) 1 4𝑥2 √1 + 𝑥4 + 𝑐

Key idea: factor $x^4$ out of the root, then the leftover is a perfect differential.

Since $\sqrt{1+x^4}=x^2\sqrt{1+x^{-4}}$, the integrand becomes

Let $u=1+x^{-4}$, so $du=-4x^{-5}\,dx$, i.e. $x^{-5}\,dx=-\tfrac14\,du$:

$\int\frac{x^{-5}\,dx}{\sqrt{1+x^{-4}}}=-\frac14\int u^{-1/2}\,du=-\frac14\cdot 2\sqrt{u}=-\frac12\sqrt{u}.$ …

We want

Why factor $x^4$ out? The derivative of $x^4$ is $4x^3$, so a bare $u=x^4$ substitution wants an $x^3$ in the numerator — but here $x^3$ sits in the denominator. Pulling $x^4$ out of the root converts the problem into one where the exact needed differential does appear.

1. Rewrite the integrand

So

2. Substitute

Let $u=1+x^{-4}$. Then $du=-4x^{-5}\,dx$, so $x^{-5}\,dx=-\tfrac14\,du$. Notice the integrand contains exactly $x^{-5}\,dx$ times $\dfrac{1}{\sqrt{u}}$:

$\int\frac{x^{-5}\,dx}{\sqrt{1+x^{-4}}}=\int\frac{-\tfrac14\,du}{\sqrt{u}}=-\frac14\int u^{-1/2}\,du.$ …

Method: Substitution when a high power of $x$ blocks the obvious $u$

Use this for integrands like $\dfrac{1}{x^{m}\sqrt{1+x^{n}}}$ where a direct substitution $u=1+x^n$ fails because the needed $x^{n-1}$ sits in the denominator, not the numerator.

Steps

Step 1: Factor the highest power of $x$ out of the root.

This deliberately introduces a negative power of $x$, which is the differential you actually need.

Step 2: Collect all powers of $x$ into one factor.

Rewrite the whole integrand so it reads (power of $x$) $\times\dfrac{1}{\sqrt{1+x^{-n}}}$. The power of $x$ should now match the derivative of $x^{-n}$.

Step 3: Substitute $u=1+x^{-n}$. …

Common Mistakes

Mistake 1: Trying $u=1+x^4$ directly.

Why it's wrong: then $du=4x^3\,dx$ needs an $x^3$ in the numerator, but here $x^3$ is in the denominator — the substitution leaves stray $x$'s. Correct approach: factor $x^4$ out of the root first to manufacture the $x^{-5}\,dx$ that $u=1+x^{-4}$ needs.

Mistake 2: Mishandling $\sqrt{x^4}=x^2$ signs.

Why it's wrong: $\sqrt{x^4}=x^2$ is fine, but pulling out $x$-powers carelessly (e.g. $\sqrt{x^4}=x$) corrupts the algebra. Correct approach: track exponents precisely — $\sqrt{x^4}=x^2$, and $\sqrt{1+x^{-4}}=\sqrt{1+x^4}/x^2$. …