Question

If \(\int f(x)\,dx = g(x)\) then \(\int x^3f(x^2)\,dx\) is equal to

• A. \(\frac{1}{2}[x^2g(x^2)-\int g(x^2)\,d(x^2)]\)
• B. \(\frac{1}{2}[x^2[f(x)]^2-\int [g(x)]^2\,dx]\)
• C. \(\frac{1}{2}[x^2g(x)-\int g(x)\,d(x)]\)
• D. \(\frac{1}{2}[x^2g(x^2)+\int g(x^2)\,d(x^2)]\)

Hint:

Split x^2/((x^2+2)(x^2+5)) into partial fractions in x^2.

Answer 1

The Correct Option is A

Step 1: Partial fractions:
Write \(\frac{x^2}{(x^2 + 2)(x^2 + 5)} = \frac{A}{x^2 + 2} + \frac{B}{x^2 + 5}\). Then \(x^2 = A(x^2 + 5) + B(x^2 + 2)\).

Step 2: Find A and B:
Compare coefficients: \(A + B = 1\) and \(5A + 2B = 0\). So \(A = -\frac23\) and \(B = \frac53\).

Step 3: Integrate:
\[ \int\frac{x^2\,dx}{(x^2+2)(x^2+5)} = -\frac23\cdot\frac{1}{\sqrt2}\tan^{-1}\frac{x}{\sqrt2} + \frac53\cdot\frac{1}{\sqrt5}\tan^{-1}\frac{x}{\sqrt5} + c \]
\[ = -\frac{\sqrt2}{3}\tan^{-1}\frac{x}{\sqrt2} + \frac{\sqrt5}{3}\tan^{-1}\frac{x}{\sqrt5} + c \]
since \(\frac{2}{3\sqrt2} = \frac{\sqrt2}{3}\) and \(\frac{5}{3\sqrt5} = \frac{\sqrt5}{3}\). The signs differ because A is negative and B is positive.

Final Answer:
The result matches option (D). \[ \boxed{-\frac{\sqrt{2}}{3}\tan^{-1}\frac{x}{\sqrt{2}} + \frac{\sqrt{5}}{3}\tan^{-1}\frac{x}{\sqrt{5}} + c} \]