What is the integral of 1/9 x+1/(9x)+(1/9)^x ?

The integral of 1/9 x+1/(9x)+(1/9)^x is (x^2)/(18)+1/9 ln|x|-1/(2ln(3)*9^x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/9 x+1/(9x)+(1/9)^x

\int \frac{1}{9} x + \frac{1}{9 x} + (\frac{1}{9})^{x} d x

\frac{x ^{2}}{18} + \frac{1}{9} ln ∣ x ∣ - \frac{1}{2 ln ( 3 ) \cdot 9 ^{x}} + C

Solution steps

\int \frac{1}{9} x + \frac{1}{9 x} + (\frac{1}{9})^{x} 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}{9} x d x + \int \frac{1}{9 x} d x + \int (\frac{1}{9})^{x} d x

\int \frac{1}{9} x d x = \frac{x ^{2}}{18}

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int (\frac{1}{9})^{x} d x = - \frac{1}{2 ln ( 3 ) \cdot 9 ^{x}}

= \frac{x ^{2}}{18} + \frac{1}{9} ln ∣ x ∣ - \frac{1}{2 ln ( 3 ) \cdot 9 ^{x}}

Add a constant to the solution

= \frac{x ^{2}}{18} + \frac{1}{9} ln ∣ x ∣ - \frac{1}{2 ln ( 3 ) \cdot 9 ^{x}} + C

\int (8 - x^{3})^{2} d x

\int \frac{6}{5 - 12 w + 9 w ^{2}} d w

\int x^{1.1} + 11 x^{4.5} d x

\int x^{2} - e^{x} d x

\int cot^{2} (t) d t

What is the integral of 1/9 x+1/(9x)+(1/9)^x ?
The integral of 1/9 x+1/(9x)+(1/9)^x is (x^2)/(18)+1/9 ln|x|-1/(2ln(3)*9^x)+C