4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Let $x=9u$. Then $dx=9\,du$. Continue. $\endgroup$ – André Nicolas Jan 7 '14 at 21:49
  • $\begingroup$ I think the problem is in computing derivative of arctan. Do you understand what its derivative is? $\endgroup$ – user114628 Jan 7 '14 at 21:52
1
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.