Integrating $\int \frac{dx}{1+\frac{a}{x}}$

Question

First-year problem:

How do I integrate

$$ \int \frac{dx}{1+\frac{a}{x}}? $$

My guess is $u$-substitution, or partial fractions, but nothing that I try seems to work...

Answer 1

\begin{align*} \int\frac{dx}{1+a/x}&=\int\frac{x\,dx}{x+a}\\[1em] &=\int\left(1-\frac{a}{x+a}\right) dx\\[1em] &=x-a\ln |x+a|+C \end{align*}

Answer 2

Hint Multiply both numerator and denominator by $x$ to rewrite the integral as $$\int \frac{x \,dx}{x + a}.$$

Answer 3

$$\int \frac{dx}{1+\frac{a}{x}}=\int \frac{x}{x+a}dx=\int \left( \frac{x+a}{x+a}-\frac{a}{x+a}\right)dx=\int \left( 1-\frac{a}{x+a}\right)dx$$

Can you continue?

Answer 4

Hint:

$$\frac1{1+\frac ax}=\frac x{x+a}=1-\frac a{x+a}$$

Answer 5

Hint: $$\int \frac{dx}{1+a/x}=\int \frac{x\,dx}{x+a}$$ and set $u=x+a$

Answer 6

$$ \int \frac{dx}{1+a/x}=\int \frac {x\,dx}{x+a}\,dx$$ Now let $u = x+a$, $du=dx$ and $x = u-a$. This gives you $$\begin{align}\int \frac{(u-a)\,du}{u} & = \int\left(1 - \frac au\right)\,du\\ & = u - a\ln|u| + C \\ &= (x+a)- a\ln|(x+a)| + C \\ &= x-a\ln|x+a|+c\end{align}$$

Answer 7

Obseve that

\begin{equation*} \int (1-\frac{a}{a+x})dx\\ =-a\int\frac{1}{u}du+\int 1dx \end{equation*}

by using the substitution $u=a+x.$ Then

\begin{equation*} x-a\log(u). \end{equation*}

Substitute back to get

\begin{equation*} x-a\log|a+x|+C. \end{equation*}

Answer 8

Simple, just consider :

$$\frac{1}{1+\frac{a}{x}} = \frac{x}{a+x}=\frac{-a + a +x}{a+x} = 1 - \frac{a}{a+x} $$

So the integral is :

\begin{eqnarray} \int \frac{1}{1+\frac{a}{x}} &=& \int dx - a\int\frac{dx}{a+x} \\ &=& x - a ln(a+x) \end{eqnarray}