Indefinite Integration
Integration Using Substitution
Grade 12
Question:
<p>\(\int \frac{e^{\tan^{-1} x}}{1+x^2}\,dx\) is equal to</p>
<p>(a) \(-e^{\tan^{-1} x} + C\)</p>
<p>(b) \(e^{\tan^{-1} x} + C\)</p>
<p>(c) \(\tan^{-1} x + C\)</p>
<p>(d) \(-\tan^{-1} x + C\)</p>
Step-by-Step Solution
Key Concept: Recognize that the denominator \(1+x^2\) is the derivative of \(\tan^{-1} x\), making this a straightforward substitution.
<p>Let \(u = \tan^{-1} x\), then \(du = \frac{1}{1+x^2}dx\). The integral becomes \(\int e^u\,du = e^u + C = e^{\tan^{-1} x} + C\).</p>
Correct Answer: B