Indefinite integral question.

Question

Evaluate the following indefinite integral.

$$ \int { \frac { 8 }{ 81+{ x }^{ 2 } } } dx $$

The answer is

$$ \frac { 8 }{ 9 } \arctan \left(\frac { x }{ 9 } \right)$$

I know that it has something to do with this integral

$$ \int { \frac { 1 }{ { x }^{ 2 }+1 } } dx =\arctan x +C $$

but i can't get it.

I think i can't make the algebra to find the antiderivative.

Answer 1

First get $\frac{8}{81}$ out of the integral (linearity) and you are left with: $\int\frac{1}{1+ (x/9)^2}dx$, now transform: $t = x/9$ so $9dt= dx$. You obtain $\frac{8}{81}\int\frac{9}{1+t^2}dt$. And now you of course obtain

$$ \begin{align} \int\frac{8}{81+x^2}dx &=\frac{8}{9} \arctan (t) + c \\ &= \frac{8}{9} \arctan (x/9) +c \end{align} $$

Answer 2

Hint: $$\int \frac{8}{81+x^2}\mathop{dx} = \frac{8}{81}\int\frac{1}{1+(x/9)^2}\mathop{dx}=\frac{8}{81}\int\frac{1}{1+u^2}9\mathop{du}$$

Answer 3

Hint: Try to transform the integrand into something resembling $\frac{1}{1+\text{something}^2}$. Use the substitution $u = \text{something}$.

Answer 4

For $$\int\frac{8}{81+x^2} \, dx,$$ If we let $$ \begin{align*} x &=9\tan \theta \\ dx&=9\sec^2 \theta \, d \theta \\ 81+x^2 &=81+81\tan^2 \theta \\ &=81 \sec^2 \theta. \end{align*} $$

We substitute,

$$ \begin{align*} \int\frac{8}{81+x^2} \, dx &=8 \int \frac{9\sec^2 \theta \, d \theta}{81\sec^2 \theta} \\ &=\frac{8}{9}\int 1 \, d \theta \\ &=\frac{8}{9} \theta +c. \end{align*} $$ For the back substitution, $$x=9\tan \theta \Rightarrow \tan \theta=\frac{x}{9} \Rightarrow \theta=\tan^{-1}\left( \frac{x}{9} \right)$$

Hence our integral, $$\frac{8}{9}\tan^{-1}\left( \frac{x}{9} \right)+c.$$