Use the substition $u=4x-3$ to find $\int\frac{4x}{4x-3}dx$, giving your answer in terms of $x$ $$u=4x-3$$ $$4x=u+3$$ $$\frac{du}{dx}=4$$ $$dx=\frac{1}{4}du$$ $$\int\frac{4x}{4x-3}dx$$ $$\Rightarrow\frac{1}{4}\int(u+3)\frac{1}{u}du$$ Integration by parts, to avoid confusion I use $a$ and $b$ $$a=u+3$$ $$\frac{da}{du}=1$$ $$\frac{db}{du}=\frac{1}{u}$$ $$b=ln(u)$$ Integration by parts formula $$\int a\frac{db}{du}du=ab-\int b\frac{da}{du}du$$ $$\Rightarrow\frac{1}{4}((u+3)ln(u)-\int ln(u))du$$ Integration by parts, again $$-\int ln(u)du$$ $$\Rightarrow-\int 1\cdot ln(u)du$$ Again, to avoid confusion I will use $y$ and $z$ $$y=ln(u)$$ $$\frac{dy}{du}=\frac{1}{u}$$ $$\frac{dz}{du}=1$$ $$z=u$$ Integration by parts formula $$\int y\frac{dz}{du}du=yz-\int z\frac{dy}{du}du$$ $$\Rightarrow u ln(u)-\int \frac{u}{u}du$$ $$\Rightarrow u ln(u)-u$$ Put every together $+c$ and substitute back in $u=4x-3$ $$\frac{1}{4}((u+3)ln(u)-\int ln(u))du$$ $$\Rightarrow\frac{1}{4}((u+3)ln(u)-(u ln(u)-u))+c$$ $$\Rightarrow\frac{1}{4}((4x)ln(4x-3)-(4x-3)ln(4x-3)+(4x-3))+c$$
The answer according to the Mark scheme is $$\frac{1}{4}((4x-3)+3ln(4x-3))+c$$
Where did I go wrong, thanks
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$\begingroup$Your answer is correct after some simplifications. But there is a much faster way: rather than integrate by parts, simply write $\frac{u+3}{u}=1+\frac{3}{u}$, then integrate each term separately.
$\endgroup$ $\begingroup$\begin{align} I&=\int \!4\,{\frac {x}{4\,x-3}}\,{\rm d}x\\ &=4\,\int \!{\frac {x}{4\,x-3}}\,{\rm d}x\\ &=4\,\int \!\bigg\{\frac{1}{4}+\frac{3}{4}\, \frac{1}{\left( 4\,x-3 \right)}\bigg\}\,{\rm d}x\\ &=4\,\int \!\frac{1}{4}\,{\rm d}x+4\,\int \!\frac{3}{4}\, \frac{1}{\left( 4\,x-3 \right)}\,{\rm d}x\\ &=x+4\,\int \!\frac{3}{4}\, \frac{1}{\left( 4\,x-3 \right)}\,{\rm d}x\\ &=x+3\,\int \! \frac{1}{\left( 4\,x-3 \right)}\,{\rm d}x\\ &=x+3\,\int \!\frac{1}{4}\,\frac{1}{u}\,{\rm d}u\\ &=x+\frac{3}{4}\,\int \!\frac{1}{u}\,{\rm d}u\\ &=x+\frac{3}{4}\,\ln \left( u \right)+C\\ &=x+\frac{3}{4}\,\ln \left( 4\,x-3 \right)+C \end{align}
$\endgroup$ $\begingroup$As others have mentioned, you have arrived at the correct answer by the long route. There were easier paths, but you got there in the end.
You had finished with: $$\frac{1}{4}((4x)\ln(4x-3)-(4x-3)\ln(4x-3)+(4x-3))+c $$
And just needed to simplify a bit more:
$$\begin{align} =~&\frac{1}{4}((4x-(4x-3))\ln(4x-3)+(4x-3))+c \\[1ex] =~&\frac{1}{4}(3\ln(4x-3)+(4x-3))+c & \text{the mark scheme's answer} \\[1ex] =~& \frac 34\ln(4x-3)+ x + (c-\frac 3 4) \\[1ex] =~& x+\frac 34\ln(4x-3) + c_1 & \text{the simplest answer} \end{align}$$
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