Question:medium

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

Show Hint

If integration is with respect to \(x^2\), put \(t=x^2\) and integrate in terms of \(t\).
  • \(x+\frac{x^3}{3}+\frac{x^5}{5}+C\)
  • \(\frac{x^2}{3}+\frac{x^4}{5}+C\)
  • \(x^2+\frac{x^4}{4}+\frac{x^6}{6}+C\)
  • \(x^2+\frac{x^4}{2}+\frac{x^6}{3}+C\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Question:
Usually, integration is performed with respect to \( x \) (the symbol \( dx \)).
However, this question specifically asks for integration "with respect to \( x^2 \)". This means the variable of integration is not \( x \) but \( u = x^2 \).
The symbol \( dx^2 \) effectively means we treat \( x^2 \) as the single variable in the power rule.
Step 2: Key Formula or Approach:
1. Let \( u = x^2 \).
2. Rewrite the function \( f(x) \) in terms of \( u \).
3. Perform integration \( \int f(u) du \).
Step 3: Detailed Explanation:

Let \( u = x^2 \). Then the function \( f(x) \) becomes:
\[ f(u) = 1 + u + u^2 \]

The question asks for the integral \( \int (1 + u + u^2) du \).

Apply the basic power rule \( \int u^n du = \frac{u^{n+1}}{n+1} \):
\[ \int (1) du = u \]
\[ \int (u) du = \frac{u^2}{2} \]
\[ \int (u^2) du = \frac{u^3}{3} \]

Summing these together with a constant \( C \):
\[ \text{Integral} = u + \frac{u^2}{2} + \frac{u^3}{3} + C \]

Now, substitute back \( u = x^2 \):
\[ \text{Result} = (x^2) + \frac{(x^2)^2}{2} + \frac{(x^2)^3}{3} + C \]
\[ = x^2 + \frac{x^4}{2} + \frac{x^6}{3} + C \]

Step 4: Final Answer:
The integral with respect to \( x^2 \) is \( x^2 + \frac{x^4}{2} + \frac{x^6}{3} + C \).
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