Integrating e^x & 1/x (Cambridge International IGCSE Further Maths)

Integrating e^x & 1/x

How do I integrate exponentials and 1/x?

• The integrals involving$e^{x}$ and$\mathrm{ln}x$ are

$\text{integral} e^{ } x d x = e^{ } x + c$

equation

where $c$ is the constant of integration

• For the linear function$(ax+b)$, where$a$ and$b$ are constants,

$\text{integral} e^{ } a x + b d x = \frac{1}{ } a e^{ } a x + b + c$

equation

• It follows from the last result that

equation

• which can be deduced using Reverse Chain Rule

• With ln, it can be useful to write the constant of integration,$c$, as a logarithm

  • using the laws of logarithms, the answer can be written as a single term
  • equationwhere$k$ is a constant
  • This is similar to the special case of differentiating$\mathrm{ln}(ax+b)$ when$b=0$

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

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

To find $f ( x )$, we need to integrate the expression for $f ' ( x ).$* *

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

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

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

Use the two results: $\int \frac{1}{a x + b} d x = \frac{1}{a} \ln \vert a x + b \vert + c$ $\int e^{a x + b} d x = \frac{ 1}{a} e^{a x + b} + c$

$f ( x ) = 3 ( 1 \frac{1}{3} \ln \vert 3 x + 2 \vert ) + - 1 e^{4 - x} + c f ( x ) = \ln \vert 3 x + 2 \vert - e^{4 - x} + c$

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

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

Simplify and solve to find $c.$* *

$\mathrm{ln}10-e^{3}=\mathrm{ln}5-e^{3}+c\,\mathrm{ln}10-\mathrm{ln}5=c$

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

$c=\mathrm{ln}\frac{10}{5}=\mathrm{ln}2$

Rewrite the full expression for $f ( x ).$* *

equation
or by combining logs, this could be written as$f ( x ) = \ln ( 2 \vert 3 x + 2 \vert ) - e^{4 - x}$