What is the integral of (ax+b)/(px+q) ?

The integral of (ax+b)/(px+q) is b 1/p ln|q+px|+a 1/(p^2)(q+px-qln|q+px|)+C

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integral of (ax+b)/(px+q)

\int \frac{a x + b}{p x + q} d x

b \frac{1}{p} ln ∣ q + p x ∣ + a \frac{1}{p ^{2}} (q + p x - q ln ∣ q + p x ∣) + C

Solution steps

\int \frac{a x + b}{p x + q} d x

Expand

\frac{b + a x}{q + p x} : \frac{b}{q + p x} + \frac{a x}{q + p x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{b}{q + p x} d x + \int \frac{a x}{q + p x} d x

\int \frac{b}{q + p x} d x = b \frac{1}{p} ln ∣ q + p x ∣

\int \frac{a x}{q + p x} d x = a \frac{1}{p ^{2}} (q + p x - q ln ∣ q + p x ∣)

= b \frac{1}{p} ln ∣ q + p x ∣ + a \frac{1}{p ^{2}} (q + p x - q ln ∣ q + p x ∣)

Add a constant to the solution

= b \frac{1}{p} ln ∣ q + p x ∣ + a \frac{1}{p ^{2}} (q + p x - q ln ∣ q + p x ∣) + C

x (\frac{d y}{d x}) - y = x

\int_{0}^{L} \frac{x}{d + x} d x

\int \frac{d t}{t ^{3} t ^{2} - 9} d x

\frac{\partial}{\partial x} ((2 x + 5 y) e^{y})

\frac{\partial}{\partial x} (x + y - x y)

What is the integral of (ax+b)/(px+q) ?
The integral of (ax+b)/(px+q) is b 1/p ln|q+px|+a 1/(p^2)(q+px-qln|q+px|)+C