Question

If $\int \frac{x^3}{\sqrt{1 + x^2}} \, dx = a(1 + x^2)\sqrt{1 + x^2} + b\sqrt{1 + x^2} + c$ (where $c$ is a constant of integration), then the value of $3ab$ is

• A. -3
• B. -1
• C. 1
• D. 3

Hint:

Calculus Tip: When an integrand contains an algebraic square root like $\sqrt{cx^2+d}$ and an odd power of $x$ outside, substituting the expression inside the root for $t^2$ (not just $t$) eliminates fractional powers and makes integration drastically simpler.

Answer 1

The Correct Option is B

Concept: Calculus - Indefinite Integration by Substitution.

Step 1: Choose an appropriate substitution. To eliminate the square root in the denominator, let $1+x^2 = t^2$. This implies that $x^2 = t^2 - 1$.

Step 2: Differentiate the substitution. Differentiate $1+x^2 = t^2$ implicitly: $2x~dx = 2t~dt$, which simplifies to $x~dx = t~dt$. Rewrite the numerator of the integral, $x^3 dx$, to utilize this: $x^3 dx = x^2(x~dx)$.

Step 3: Substitute and simplify the integral. Substitute $t$ into the original integral: $\int \frac{x^2 \cdot x~dx}{\sqrt{1+x^2}} = \int \frac{(t^2 - 1) \cdot t~dt}{\sqrt{t^2}} = \int \frac{(t^2 - 1) \cdot t~dt}{t}$. The $t$ in the numerator and denominator cancel out perfectly, leaving a simple polynomial integral: $\int (t^2 - 1) dt$.

Step 4: Integrate and substitute $x$ back. Integrate with respect to $t$: $\frac{t^3}{3} - t + c$. Now, substitute $t = \sqrt{1+x^2}$ back into the expression: $\frac{(\sqrt{1+x^2})^3}{3} - \sqrt{1+x^2} + c$. We can write $(\sqrt{1+x^2})^3$ as $(1+x^2)\sqrt{1+x^2}$. So the expression becomes: $\frac{1}{3}(1+x^2)\sqrt{1+x^2} - 1\sqrt{1+x^2} + c$.

Step 5: Compare coefficients and calculate $3ab$. Compare our integrated result with the given expression: $a(1+x^2)\sqrt{1+x^2} + b\sqrt{1+x^2} + c$. By matching coefficients, we find that $a = \frac{1}{3}$ and $b = -1$. The question asks for the value of $3ab$. Substitute the values: $3 \cdot \left(\frac{1}{3}\right) \cdot (-1) = 1 \cdot (-1) = -1$. $$ \therefore \text{The value of } 3ab \text{ is } -1. $$