Question

The integral of \(f(x)=1+x^2+x^4\) with respect to \(x^2\) is

• A. \(x+\frac{x^3}{3}+\frac{x^5}{5}+C\)
• B. \(\frac{x^2}{3}+\frac{x^4}{5}+C\)
• C. \(x^2+\frac{x^4}{4}+\frac{x^6}{6}+C\)
• D. \(x^2+\frac{x^4}{2}+\frac{x^6}{3}+C\)

Hint:

If integration is with respect to \(x^2\), put \(t=x^2\) and integrate in terms of \(t\).

Answer 1

The Correct Option is D

Concept: The phrase “with respect to \(x^2\)” means we should treat \(x^2\) as the variable of integration.

Step 1: Let: \[ t=x^2 \] Then: \[ x^4=(x^2)^2=t^2 \]

Step 2: Rewrite the function in terms of \(t\). \[ f(x)=1+x^2+x^4 \] \[ f(t)=1+t+t^2 \]

Step 3: Now integrate with respect to \(t\). \[ \int (1+t+t^2)\,dt \] \[ = t+\frac{t^2}{2}+\frac{t^3}{3}+C \]

Step 4: Substitute \(t=x^2\). \[ = x^2+\frac{(x^2)^2}{2}+\frac{(x^2)^3}{3}+C \] \[ = x^2+\frac{x^4}{2}+\frac{x^6}{3}+C \] Therefore, \[ \boxed{x^2+\frac{x^4}{2}+\frac{x^6}{3}+C} \]