1. ## Integration help

Could someone assist on the two problems below?

1. Find the anti-derivative by parts:

ln3x/x^2 I get the following answer: -1/x*ln3x-1/2x^2 + C

2. Solve by substitution:
x*(x-3)^1/2 If I use u=(x-3)^1/2 I don't know how to get rid of the first x

Any help will be appreciated.

2. Originally Posted by Wildcatz
Could someone assist on the two problems below?

1. Find the anti-derivative by parts:

ln3x/x^2 I get the following answer: -1/x*ln3x-1/2x^2 + C
this is not correct. try again. what part did you decide to differentiate? which did you integrate?

3. Originally Posted by Wildcatz
2. Solve by substitution:
x*(x-3)^1/2 If I use u=(x-3)^1/2 I don't know how to get rid of the first x

Any help will be appreciated.
wrong substitution. let u = x - 3

4. Let $u=ln(3x)$ and $dv=x^{-2}\:dx$

Thus, $du=\dfrac{1}{x}\:dx$ and $v=-\dfrac{1}{x}$

$\int \dfrac{ln(3x)}{x^2} = -\dfrac{ln(3x)}{x}+\int\dfrac{1}{x^2}$
$\int \dfrac{ln(3x)}{x^2} = -\dfrac{ln(3x)}{x} + ({-\dfrac{1}{x}})$

Therefore, $\int \dfrac{ln(3x)}{x^2} = -(\dfrac{ln(3x)+1}{x})$

5. $\int{x\cdot{\sqrt[]{x-3}}}dx=2\cdot{\int{\frac{x\cdot{(x-3)}}{2\cdot{\sqrt[]{x-3}}}}dx}$

Let $u=\sqrt[]{x-3}$ thus: $\frac{du}{dx}=\frac{1}{2\cdot{\sqrt[]{x-3}}}$

Then: $\int{x\cdot{\sqrt[]{x-3}}}dx=2\cdot{\int{(u^2+3)\cdot{u^2}}du}$

$2\cdot{\int{(u^2+3)\cdot{u^2}}du}=2\cdot{\left(\fr ac{u^5}{5}+u^3\right)}+k$

So $\int{x\cdot{\sqrt[]{x-3}}}dx=2\cdot{\left(\frac{(\sqrt[]{x-3})^5}{5}+(\sqrt[]{x-3})^3\right)}+k$

Which is: $\int{x\cdot{\sqrt[]{x-3}}}dx=\frac{2}{5}\cdot{(x-3)^{3/2}\cdot{(x+2)}}+k$

6. Originally Posted by Wildcatz
2. Solve by substitution:
x*(x-3)^1/2 If I use u=(x-3)^1/2 I don't know how to get rid of the first x

Originally Posted by Jhevon
wrong substitution. let u = x - 3
I think it's quickly substitution, this leads to (by setting $u^2=x-3),$

$\int {x\sqrt {x - 3} \,dx} = 2\int {u^2 \left( {u^2 + 3} \right)\,du} .$

The conclusion follows from PaulRS' evaluation.