# Question about Calculus II: Integration by Parts

I'm trying to figure out my mistakes in my calculations, I know it's wrong because I don't get the right answer. I'm tying to integrate $7x\ln(6x)$ using integration by parts.

My answer is$\frac{42}{72}\ln(6x)x^2 - \frac{42}{144}x^2 + C$ However, the right answer is :$\frac72 \ln(6x)x^2 - \frac74x^2 + C$

I substituted $3x$ into $u$.

Can anyone explain?

Thanks.

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How did you get $72$ and $144$? – Berci Mar 11 '13 at 1:14
It's much harder to help you figure out the mistakes in your calculations without seeing your calculations. – mixedmath Mar 11 '13 at 1:16
I second @Berci's comment. Where did you get these numbers? – copper.hat Mar 11 '13 at 1:22

$$(fg)'=f'g+fg' \\fg =\int f'g+\int fg'$$ Apply it to an $f$ with $f'=7x$ and $g=\ln(6x)$. Well, $f$ then can be $\frac72x^2$. And you also need $g'(x)=\frac6{6x}=\frac1x$.

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We have $u=\ln6x$, then $du=\frac{1}{x^{2}}dx$. Also, $dv=7x dx$, then $v=\frac{7x^2}{2}$.

Integration by parts gives you

$\int 7x\ln6x dx=uv-\int vdu= \ln6x\frac{7x^2}{2}-\int\frac{7x^2}{2}\frac{1}{x^{2}}dx = \ln6x\frac{7x^2}{2}- \frac{7x^2}{4} + C$.

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$u = \ln (6x)$, $dv = 7x dx$ gives $du = \frac{1}{x} dx$, $v = \frac{7}{2} x^2$.

$\int u dv = uv-\int v du = \frac{7}{2} x^2 \ln (6x) - \int \frac{7}{2} x^2 \frac{1}{x} dx = \frac{7}{2} x^2 \ln (6x) -\frac{7}{4}x^2$

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