Q.If $\int \dfrac{x^3}{\sqrt{1+x^2}}\,dx = a(1+x^2)^{3/2} + b\sqrt{1+x^2} + C$, then
(A) $a=\dfrac{1}{3},\ b=1$
(B) $a=-\dfrac{1}{3},\ b=1$
(C) $a=-\dfrac{1}{3},\ b=-1$
(D) $a=\dfrac{1}{3},\ b=-1$

Concept: Integration by substitution — rewrite $x^3 = x \cdot x^2$ and use $1+x^2$ as the base.

Step 1: Let $t = 1+x^2$, so $dt = 2x\,dx$ and $x^2 = t-1$. Then

Step 2: Simplify:

$\frac12 \int \left( t^{1/2} - t^{-1/2} \right) dt = \frac12 \left( \frac{2}{3} t^{3/2} - 2 t^{1/2} \right) + C = \frac13 t^{3/2} - t^{1/2} + C.$ …

The problem gives us a definite structure for the antiderivative:

$\int \frac{x^3}{\sqrt{1+x^2}}\,dx = a(1+x^2)^{3/2} + b\sqrt{1+x^2} + C$.

We need to find $a$ and $b$. The direct approach is to actually compute the integral and then match coefficients.

Why "completing the square" in the numerator?

The denominator is $\sqrt{1+x^2}$, which suggests that a substitution $t = 1+x^2$ will simplify the square root. But the numerator $x^3$ is not a simple multiple of $dx$ in that substitution — unless we rewrite it cleverly.

Notice:

If $t = 1+x^2$, then $dt = 2x\,dx$. So we need an $x\,dx$ factor. The numerator $x^3$ can be split as $x^2 \cdot x$, and $x^2$ itself can be expressed in terms of $t$: $x^2 = t-1$. This is the "completing the square" idea — not literally completing a square, but rewriting the integrand so that the substitution works cleanly.

Step-by-step computation

1. Rewrite the integrand

We have:

2. Substitute $t = 1+x^2$

Then $dt = 2x\,dx$, so $x\,dx = \frac{dt}{2}$.

Also $x^2 = t-1$.

The integral becomes:

3. Simplify the integrand

So the integral is:

4. Integrate using the power rule

Simplify: …

Method: Splitting an odd power for a $t=1+x^2$ substitution

For $\int\dfrac{x^{2k+1}}{\sqrt{1+x^2}}\,dx$, peel off one $x$ to pair with $dt$ and convert the remaining even power via $x^2=t-1$.

Steps

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

$dt=2x\,dx$ and $x^2=t-1$. Write $x^3=x^2\cdot x$, so $x^3\,dx=(t-1)\cdot\tfrac{dt}{2}$.

Step 2: Simplify the powers of $t$. …

Common Mistakes

Mistake 1: Choosing $b=+1$ (option A).

Why it's wrong: the integral yields $-\sqrt{1+x^2}$, so $b=-1$. Correct approach: track the $-t^{1/2}$ term through the back-substitution.

Mistake 2: Not converting the leftover even power with $x^2=t-1$. …