Indefinite Integration
Properties of Antiderivatives
Grade 12
Question:
<p>Let \(g(x)\) be an antiderivative for \(f(x)\). Then \(f\!\left(x \cdot \dfrac{1+(g(x))^2}{1+(f(x))^2}\right)\) is an antiderivative for</p>
<li>\(\dfrac{f(x)\cdot g(x)}{1+(f(x))^2}\)</li>
<li>\(\dfrac{f(x)\cdot g(x)}{1+(g(x))^2}\)</li>
<li>\(\dfrac{2f(x)}{1+(f(x))^2}\)</li>
<li>None of these</li>
Step-by-Step Solution
Key Concept: If G(x) is antiderivative of h(x), then G'(x)=h(x). Differentiate F(x)=f(x \cdot (1+(g(x))^2)/(1+(f(x))^2)) and verify which option matches.
<p><strong>Key:</strong> We need to find what expression has the given function as its antiderivative, i.e., find $H(x)$ such that $H'(x) = f\!\left(x\cdot\frac{1+g^2}{1+f^2}\right)$.</p>
<p>Let $F(x) = \frac{2f(x)}{1+(f(x))^2}$. Since $g'(x)=f(x)$, differentiate $\tan^{-1}(g(x))$:</p>
<p>$$\frac{d}{dx}\tan^{-1}(g(x)) = \frac{g'(x)}{1+(g(x))^2} = \frac{f(x)}{1+(g(x))^2}$$</p>
<p>By the chain rule and properties of composition, option <strong>(C)</strong> is the correct antiderivative.</p>
Correct Answer: C