anti-derivatives

**JGaraffa** · May 4th 2010, 11:08 AM

sorry for multiple posts today, i'm studying for my calc final and just wanna make sure i know the correct order of solving these.

"solve the anti-derivative"
$<br /> f(x) = \frac{1}{x}(lnx + 3)$

do i solve them separately? or distribute the $\frac{1}{x}$ ?

**Failure** · May 4th 2010, 11:13 AM

Originally Posted by JGaraffa

sorry for multiple posts today, i'm studying for my calc final and just wanna make sure i know the correct order of solving these.

"solve the anti-derivative"
$<br /> f(x) = \frac{1}{x}(lnx + 3)$

do i solve them separately? or distribute the $\frac{1}{x}$ ?

Well, again, use substitution $z := \ln x+3$ , which gives you that $dz = \tfrac{1}{x}\,dx$ and hence

$\int \tfrac{1}{x}(\ln x+3)\, dx=\int z\, dz=\ldots$

**JGaraffa** · May 4th 2010, 11:36 AM

got it, thanks man.

i just posted a reply in the other thread "definite integral" which you responded to, i'm still a little confused if i'm solving it right (the numbers came out really high) if you could just check it out that would be great.

**CaptainBlack** · May 4th 2010, 11:24 PM

Originally Posted by JGaraffa

sorry for multiple posts today, i'm studying for my calc final and just wanna make sure i know the correct order of solving these.

"solve the anti-derivative"
$<br /> f(x) = \frac{1}{x}(lnx + 3)$

do i solve them separately? or distribute the $\frac{1}{x}$ ?

Again observe that $1/x$ is the derivative of $\ln(x)+3$ so what is the derivative of:

$(\ln(x)+3)^2$ ?

CB

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