Integrating e^x & 1/x (Cambridge (CIE) IGCSE Additional Maths): Revision Note

A curve has the gradient function$f\text{'}\left(x\right)=\frac{3}{3x+2}+{\mathrm{e}}^{4-x}$.

Given the exact value of$f\left(1\right)$ is$\ln 10-{\mathrm{e}}^3$ find an expression for$f\left(x\right)$.

To find $f\left(x\right)$, we need to integrate the expression for $f\text{'}\left(x\right).$   

$f\left(x\right)=\int \frac{3}{3x+2}+e^{4-x}dx$  

Rewrite as two separate integrals which can be found individually, and factor out any constants.  

$f\left(x\right)=3\int \frac{1}{3x+2}dx+\int e^{4-x}dx$  

Use the two results: $\int \frac{1}{ax+b}dx=\frac{1}{a}\ln |ax+b|+c$ $\int e^{ax+b}dx=\frac{1}{a}{e}^{ax+b}+c$   

$\begin{array}{l}f\left(x\right)=3\left(\frac{1}{3}\ln |3x+2|\right)+-1e^{4-x}+c\\ f\left(x\right)=\ln |3x+2|-e^{4-x}+c\end{array}$  

The question states that $f\left(1\right)=\ln 10-e^3$, so substitute in $x=1$ and equate to the given expression.  

$\ln 10-e^3=\ln |3\left(1\right)+2|-e^{4-1}+c$  

Simplify and solve to find $c.$  

$\begin{array}{l}\ln 10-e^3=\ln 5-e^3+c\\ \ln 10-\ln 5=c\end{array}$  

Recall from laws of logarithms that $\ln a-\ln b=\ln \frac{a}{b}.$  

$c=\ln \frac{10}{5}=\ln 2$  

Rewrite the full expression for $f\left(x\right).$  

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{ln}\mathbf{|}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{|}\mathbf{-}{\mathit{e}}^{\mathbf{4}\mathbf{-}\mathit{x}}\mathbf{+}\mathbf{ln}\mathbf{2}$ 
or by combining logs, this could be written as $f\left(x\right)=\ln \left(2|3x+2|\right)-e^{4-x}$