Question

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

• A. $-\frac{2}{3}$
• B. $-\frac{1}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Hint:

Whenever you see $x^2$ inside a root and an odd power of $x$ outside (like $x^3$ or $x^5$), substituting the entire term inside the root as $t^2$ (instead of just $t$) makes the algebra much cleaner by avoiding fractional exponents during integration!

Answer 1

The Correct Option is A

Step 1: Pick a substitution.
The square root $\sqrt{1+x^2}$ is the messy part, so let $t = \sqrt{1+x^2}$. Then $t^2 = 1+x^2$ and $x\,dx = t\,dt$. Also $x^2 = t^2 - 1$.

Step 2: Rewrite the integral.
Split $x^3$ as $x^2 \cdot x$: \[ \int \frac{x^2 \cdot x}{\sqrt{1+x^2}}\,dx = \int \frac{(t^2-1)\,t\,dt}{t} = \int (t^2-1)\,dt. \]

Step 3: Integrate and go back.
This gives $\tfrac{t^3}{3} - t + c$. Put $t = (1+x^2)^{1/2}$: \[ \tfrac{1}{3}(1+x^2)^{3/2} - \sqrt{1+x^2} + c. \]

Step 4: Read off $a$ and $b$.
So $a = \tfrac{1}{3}$ and $b = -1$, which gives $a+b = \tfrac{1}{3} - 1 = -\tfrac{2}{3}$. \[ \boxed{a+b = -\tfrac{2}{3}} \]