Q.$\int \dfrac{x^9}{(4x^2+1)^6}\,dx$ is equal to
(A) $\dfrac{1}{5x}\left(\dfrac{1}{x^2}+4\right)^{-5} + C$
(B) $\dfrac{1}{5}\left(\dfrac{1}{x^2}+4\right)^{-5} + C$
(C) $\dfrac{1}{10x}\left(\dfrac{1}{x^2}+4\right)^{-5} + C$
(D) $\dfrac{1}{10}\left(\dfrac{1}{x^2}+4\right)^{-5} + C$

Concept: U Substitution — rewrite the integral so the denominator’s structure becomes a simple power.

Step 1: Let $u = \frac{1}{x^2} + 4$. Then $du = -\frac{2}{x^3}\,dx$, so $dx = -\frac{x^3}{2}\,du$.

Step 2: Rewrite the integrand.

$\frac{x^9}{(4x^2+1)^6} = \frac{x^9}{x^{12}\left(4 + \frac{1}{x^2}\right)^6} = \frac{1}{x^3\left(\frac{1}{x^2}+4\right)^6}$.

Step 3: Substitute $u$ and $dx$: …

The first thing to notice is the high power of $x$ in the numerator ($x^9$) and the quadratic in the denominator raised to the 6th power. A direct substitution like $u = 4x^2+1$ would give $du = 8x\,dx$, but we have $x^9$, not just $x$. That mismatch suggests we need to rewrite the integrand in terms of $1/x^2$ — a classic trick when powers are heavily unbalanced.

Observe that the denominator is $(4x^2+1)^6 = x^{12}\left(4 + \frac{1}{x^2}\right)^6$. Why $x^{12}$? Because $(x^2)^6 = x^{12}$. This lets us factor $x^{12}$ out of the denominator, and then the $x^9$ in the numerator cancels partially:

Now the integrand is $\frac{1}{x^3} \cdot \left(4 + \frac{1}{x^2}\right)^{-6}$. This suggests the substitution $t = \frac{1}{x^2}$, because $dt = -\frac{2}{x^3}\,dx$, and we have exactly $\frac{1}{x^3}\,dx$ sitting there (up to a constant factor).

Let’s work through it step by step.

1. Rewrite the integrand

2. Choose the substitution

   Let $t = \frac{1}{x^2}$. Then $dt = -\frac{2}{x^3}\,dx$, so $\frac{1}{x^3}\,dx = -\frac{1}{2}\,dt$.

3. Express the integrand in terms of $t$

   The term $4 + \frac{1}{x^2}$ becomes $4 + t$. So

4. Integrate with respect to $t$

Therefore, …

Method: Factoring out $x^{2n}$ for a $t=1/x^2$ substitution

When the numerator's power is far higher than a simple substitution allows, factor a power of $x$ out of the denominator to create a $1/x^k$ that pairs with $dt$.

Steps

Step 1: Factor the denominator.

$(ax^2+c)^n=x^{2n}\left(a+\dfrac{c}{x^2}\right)^n$; cancel against the numerator's $x$-power to leave $\dfrac{1}{x^m}\left(a+\dfrac{1}{x^2}\right)^{-n}$.

Step 2: Substitute $t=1/x^2$. …

Common Mistakes

Mistake 1: Substituting $u=4x^2+1$ directly.

Why it's wrong: $du=8x\,dx$ cannot absorb the $x^9$ numerator. Correct approach: factor $x^{12}$ from the denominator and substitute $t=\frac{1}{x^2}$ instead.

Mistake 2: Mishandling the sign/half-factor from $dt$. …