What is the integral of (a^2+b)/(b-ax) ?

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

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integral of (a^2+b)/(b-ax)

\int \frac{a ^{2} + b}{b - a x} d x

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

Solution steps

\int \frac{a ^{2} + b}{b - a x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= (a^{2} + b) \cdot \int \frac{1}{b - a x} d x

\int \frac{1}{b - a x} d x = - \frac{1}{a} ln ∣ b - a x ∣

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

Simplify

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

Add a constant to the solution

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

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\int \frac{4 x ^{3}}{1 + x ^{8}} d x

\int sin (\frac{x ^{2}}{2}) d x

\int 2 x 9 + x^{2} d x

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