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: Set up the integral.
We want $\int\frac{x^3}{\sqrt{1+x^2}}\,dx$ and then match it to $a(1+x^2)^{3/2}+b\sqrt{1+x^2}+c$.
Step 2: Choose a neat substitution.
Let $t=1+x^2$. Then $dt=2x\,dx$, so $x\,dx=\tfrac{1}{2}dt$. Also $x^2=t-1$.
Step 3: Rewrite the top.
Split $x^3\,dx$ as $x^2\cdot x\,dx=(t-1)\cdot\tfrac{1}{2}dt$. The integral becomes $\int\frac{(t-1)}{\sqrt{t}}\cdot\tfrac{1}{2}\,dt$.
Step 4: Simplify the powers.
This is $\tfrac{1}{2}\int\left(t^{1/2}-t^{-1/2}\right)dt$, because $(t-1)/\sqrt{t}=\sqrt{t}-1/\sqrt{t}$.
Step 5: Integrate term by term.
We get $\tfrac{1}{2}\left(\tfrac{2}{3}t^{3/2}-2t^{1/2}\right)+c=\tfrac{1}{3}t^{3/2}-t^{1/2}+c$. Put back $t=1+x^2$: $\tfrac{1}{3}(1+x^2)^{3/2}-\sqrt{1+x^2}+c$.
Step 6: Read off and add.
Matching gives $a=\tfrac{1}{3}$ and $b=-1$, so $a+b=\tfrac{1}{3}-1=-\tfrac{2}{3}$. This is option (A).
\[ \boxed{\,a+b=-\tfrac{2}{3}\,} \]