1
$\begingroup$

How can I resolve this indefinite integral?

$$ \int \left({8,387x+1 \over 9,41x+1} + \sin(9,326x + 1)\right) dx $$

I'm blocked here

$$ \int\left({{8387 \over 1000}x+1}\over{{941\over100}x+1} \right)dx + \int\left(\sin\left({9326\over1000}x+1\right) \right)dx $$

please anyone can help me? thanks

$\endgroup$
  • $\begingroup$ Replace those strange numbers with some friendly ones so you do not get distracted while doing and learning the substitution method. Then apply what you learned to the general case. $\endgroup$ – Maesumi Jan 17 '13 at 13:47
4
$\begingroup$

As said above, I also think you are getting distracted by the "strange" numbers.

Assume you have

$\displaystyle\int\dfrac{ax+1}{bx+1}dx+\displaystyle\int\sin(cx+1)dx$,

where $a,b,c$ are some constants, then you can write

$\displaystyle\int\dfrac{ax}{bx+1}dx+\displaystyle\int\dfrac{1}{bx+1}dx+\displaystyle\int\sin(cx+1)dx$,

now just proceed by rewriting the first term as $\dfrac{a}{b}-\dfrac{a/b}{bx+1}$, this is, now you have

$\displaystyle\int\left( \dfrac{a}{b}-\dfrac{a/b}{bx+1} \right)dx+\displaystyle\int\dfrac{1}{bx+1}dx+\displaystyle\int\sin(cx+1)dx$,

and now you are able to integrate everything by declaring new variables as already suggested: $u=bx+1$ and $v=cx+1$.

$\endgroup$
1
$\begingroup$

Do a substitution $w = 9.41 \, x + 1$ for the first and $y = 9.236 \, x + 1$ for the second integral.

$\endgroup$
  • $\begingroup$ but how can i resolve the 8,387x+1? $\endgroup$ – Sam Jan 17 '13 at 12:39

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.