Question

Find the value of \(\int \frac{1}{16x^2+16x+10}dx\).

(a) ^1⁄8 sin^-1(x + ^1⁄2)

(b) ^1⁄8 tan^-1(x + ^1⁄2)

(c) ^1⁄8 sec^-1(x + ^1⁄2)

(d) ^1⁄4 cos^-1(x + ^1⁄2)

This question was addressed to me during an interview.

I want to ask this question from Improper Integrals topic in chapter Integral Calculus of Engineering Mathematics

Answer 1

Correct option is (b) ^1⁄8 tan^-1(x + ^1⁄2)

The best explanation: Add constant automatically

Given, \(\int \frac{1}{16x^2+16x+10}dx=\frac{1}{2}\int \frac{1}{4x^2+4x+5}dx\)

=\(\int \frac{1}{8(x^2+x+\frac{5}{4}+\frac{1}{4}+\frac{1}{4})}dx=\int \frac{1}{8[(x+\frac{1}{2})^2+1^2]}dx=\frac{1}{8}tan^{-1}(x+\frac{1}{2})\)