Question

Evaluate: $\int \frac{1}{9x^2+49}dx$

Answer 1

$I=\int \frac{1}{9\left(x^2\right)+49}dx$

$I=\frac{1}{9}\int \frac{1}{{\left(x\right)}^2+\left(49/9\right)}dx$

$I=\frac{1}{9}\int \frac{1}{{\left(x\right)}^2+{\left(7/3\right)}^2}dx$

We know $\int \frac{1}{{\left(x\right)}^2+a^2}dx=\frac{1}{9}{\tan }^{-1}\left(x/a\right)+c$

$\therefore I=\frac{1}{9}\left\{\frac{1}{\left(7/3\right)}{\tan }^{-1}\left(\frac{3x}{7}\right)\right\}+c$

or $I=\frac{1}{9.\frac{7}{3}}{\tan }^{-1}\left(\frac{3x}{7}\right)+c$

$I=\frac{1}{21}{\tan }^{-1}\left(\frac{3x}{7}\right)+c$