Question

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

(a) \(\frac{1}{8} (x+\frac{1}{2})Sin^{-1} (x+\frac{1}{2})-\frac{1}{2} ln⁡(1+(x+\frac{1}{2})^2)\)

(b) \(\frac{1}{8} (x+\frac{1}{2})tan^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)

(c) \((x+1/2) cos^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)

(d) \(\frac{1}{8} (x+\frac{1}{2}) sec^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)

This question was addressed to me by my college director while I was bunking the class.

Query is from Double Integrals in section Multiple Integrals of Engineering Mathematics

Answer 1

Correct choice is (b) \(\frac{1}{8} (x+\frac{1}{2})tan^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)

Best explanation: Add constant automatically

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

=\(\int \frac{1}{8(x^2+x+5/4+1/4+1/4)} dx\)

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

Hence, \(\frac{1}{8} \int tan^{-1}⁡(x+\frac{1}{2})dx\)

Now, Putting, x+1/2 = tan(y),

We get, dx = sec^2⁡(y)dy,

=1/8 \(\int ysec^2 (y)dy\)

By integration by parts,

ytan(y)-log⁡(sec⁡(y))=\(\frac{1}{8} (x+\frac{1}{2})tan^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)