Indefinite Integration Question - None

Question

If $x^2 d(\tan^{-1} x) = x f(x) + c$, then $f(1)$ is equal to

Accepted Answer

Step 1: Start with $x^2 d( an^{-1} x) = x f(x) + c$

Step 2: We know $d( an^{-1} x) = \frac{1}{1+x^2} dx$

Step 3: So $x^2 \cdot \frac{1}{1+x^2} dx = x f(x) + c$

Step 4: Integrating: $\int \frac{x^2}{1+x^2} dx = \int f(x) dx + c$

Step 5: $\int \left(1 - \frac{1}{1+x^2} ight) dx = \int f(x) dx$

Step 6: $x - an^{-1} x = \int f(x) dx$

Step 7: Therefore $f(x) = 1 - \frac{1}{1+x^2}$

Step 8: At $x=1$: $f(1) = 1 - \frac{1}{2} = \frac{1}{2}$

Note: If the matching gives P as 0, this may require rechecking the original problem statement. Based on standard integration, the answer is $\frac{1}{2}$.

Question:

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