What is the integral of (1+ax^b)/(x^b) ?

The integral of (1+ax^b)/(x^b) is (x^{-b+1})/(-b+1)+ax+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1+ax^b)/(x^b)

\int \frac{1 + a x ^{b}}{x ^{b}} d x

\frac{x ^{- b + 1}}{- b + 1} + a x + C

Solution steps

\int \frac{1 + a x ^{b}}{x ^{b}} d x

Expand

\frac{1 + a x ^{b}}{x ^{b}} : \frac{1}{x ^{b}} + a

= \int \frac{1}{x ^{b}} + a d 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{1}{x ^{b}} d x + \int a d x

\int \frac{1}{x ^{b}} d x = \frac{x ^{- b + 1}}{- b + 1}

\int a d x = a x

= \frac{x ^{- b + 1}}{- b + 1} + a x

Add a constant to the solution

= \frac{x ^{- b + 1}}{- b + 1} + a x + C

\int \frac{1}{x ^{2} + 4 x - 5} d x

\int - x^{2} + 3 x - 2 d x

\int \frac{2 x + 3}{x ^{2} + 1} d x

\int 8 x e^{4 x} d x

\int \frac{3}{3 x + 5} d x

What is the integral of (1+ax^b)/(x^b) ?
The integral of (1+ax^b)/(x^b) is (x^{-b+1})/(-b+1)+ax+C