Question

Integrate the function : $\frac{1}{9x^2 + 6x+5}$

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Solution:

$9 {x}^{2} + 6x + 5 \\ \\ {(3x)}^{2} + 2(3x)(1) + ( {1)}^{2} + 4 \\ \\9 {x}^{2} + 6x + 5 = {(3x + 1)}^{2} + 4$
$\int\frac{1}{9x^2 + 6x+5}dx \\ \\ = \int\frac{1}{{(3x + 1)}^{2} + 4}dx \\ \\ = \frac{1}{4} \int \frac{1}{ { (\frac{3x + 1}{2} })^{2} + 1} dx \\ \\ let \: \frac{3x + 1}{2} = t \\ \frac{3}{2} dx = dt \\ \\ dx = \frac{2}{3} dt$
after substitution

$\frac{1}{4} \times \frac{2}{3} \int \frac{1}{ {t}^{2} + 1 } dt \\ \\ = \frac{1}{6} {tan}^{ - 1} t + C$
redo substitution

$\int\frac{1}{9x^2 + 6x+5}dx=\frac{1}{6} {tan}^{ - 1} \bigg( \frac{3x + 1}{2}\bigg) + C$
Hope it helps you.