Integration of (ax+b)/(cx+d): Formula, Proof

The integration of (ax+b)/(cx+d) is equal to ax/c + (bc-ad)/c² ln|cx+d|+K where K is an integral constant and ln denotes the natural logarithm. In this post, we will learn how to integrate the function (ax+b)/(cx+d).

Integration of (ax+b)/(cx+d) Formula

The integration of (ax+b)/(cx+d) formula is given below.

$\int \dfrac{ax+b}{cx+d} \ dx$ $=\dfrac{ax}{c} + \dfrac{bc-ad}{c^2} \ln|cx+d|+K$,

where K is an integration constant.

How to Integrate (ax+b)/(cx+d)

To find the integration of (ax+b)/(cx+d), we will proceed as follows.

$I=\int \dfrac{ax+b}{cx+d} \ dx$

= $\int \Big(\dfrac{ax}{cx+d}+\dfrac{b}{cx+d}\Big) \ dx$

= $\int \dfrac{ax}{cx+d} \ dx$ $+ \int \dfrac{b}{cx+d} \ dx$

= $\dfrac{a}{c}\int \dfrac{cx}{cx+d} \ dx$ $+\dfrac{b}{c} \int \dfrac{c}{cx+d} \ dx$

= $\dfrac{a}{c}\int \dfrac{cx+d-d}{cx+d} \ dx$ $+\dfrac{b}{c} \int \dfrac{c}{cx+d} \ dx$

= $\dfrac{a}{c}\int \Big(\dfrac{cx+d}{cx+d} -\dfrac{d}{cx+d} \Big)\ dx$ $+\dfrac{b}{c} \int \dfrac{c}{cx+d} \ dx$

= $\dfrac{a}{c}\int dx – \dfrac{ad}{c^2} \int \dfrac{c}{cx+d} \ dx$ $+\dfrac{b}{c} \int \dfrac{c}{cx+d} \ dx$

= $\dfrac{a}{c}\int dx$ $+ \Big(\dfrac{b}{c}-\dfrac{ad}{c^2}\Big) \int \dfrac{c}{cx+d} \ dx$

Let cx+d=z.

Differentiating both sides, c dx=dz.

So from above, we obtain that

$I=\dfrac{a}{c} \times x + \dfrac{bc-ad}{c^2} \int \dfrac{dz}{z}+C$

= $\dfrac{ax}{c} + \dfrac{bc-ad}{c^2} \ln|z|+C$

= $\dfrac{ax}{c} + \dfrac{bc-ad}{c^2} \ln|cx+d|+C$ as z=cx+d.

So the integration of (ax+b)/(cx+d) is equal to $\dfrac{ax}{c} + \dfrac{bc-ad}{c^2} \ln|cx+d|+C$ where C is an integral constant.

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