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: Understanding the Question:
We need to evaluate an indefinite integral, express the result in a specific algebraic form with coefficients $a$ and $b$, and then calculate the sum of these coefficients.

Step 2: Key Formula or Approach:
Use the method of substitution to eliminate the square root in the denominator.
Let $1 + x^2 = t^2$. Differentiating both sides gives $2x \, dx = 2t \, dt$, which simplifies to $x \, dx = t \, dt$.
We will also need to split $x^3$ in the numerator into $x^2 \cdot x$ so we can substitute $x^2 = t^2 - 1$.

Step 3: Detailed Explanation:
Let the integral be $I$:
$$I = \int \frac{x^3}{\sqrt{1+x^2}} \, dx$$ Rewrite $x^3$ as $x^2 \cdot x$:
$$I = \int \frac{x^2 \cdot x}{\sqrt{1+x^2}} \, dx$$ Substitute $1+x^2 = t^2 \implies \sqrt{1+x^2} = t$.
Substitute $x^2 = t^2 - 1$.
Substitute $x \, dx = t \, dt$.
$$I = \int \frac{(t^2 - 1) \cdot t \, dt}{t}$$ The $t$ in the numerator and denominator cancel out perfectly:
$$I = \int (t^2 - 1) \, dt$$ Integrate with respect to $t$:
$$I = \frac{t^3}{3} - t + c$$ Now, substitute back $t = (1+x^2)^{\frac{1}{2}}$:
$$I = \frac{((1+x^2)^{\frac{1}{2}})^3}{3} - (1+x^2)^{\frac{1}{2}} + c$$ $$I = \frac{1}{3}(1+x^2)^{\frac{3}{2}} - \sqrt{1+x^2} + c$$ Compare this result with the given expression: $a(1+x^2)^{\frac{3}{2}} + b\sqrt{1+x^2} + c$.
By matching the coefficients, we find:
$$a = \frac{1}{3}$$ $$b = -1$$ Calculate the required sum:
$$a + b = \frac{1}{3} + (-1) = \frac{1}{3} - \frac{3}{3} = -\frac{2}{3}$$

Step 4: Final Answer:
The value of $a+b$ is $-\frac{2}{3}$, which corresponds to option (A).