What is the integral of (x^{1.9}+9x^{3.5}) ?

The integral of (x^{1.9}+9x^{3.5}) is 0.34482…x^{2.9}+2x^{4.5}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^{1.9}+9x^{3.5})

\int (x^{1.9} + 9 x^{3.5}) d x

0.34482 \dots x^{2.9} + 2 x^{4.5} + C

Solution steps

\int x^{1.9} + 9 x^{3.5} 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 x^{1.9} d x + \int 9 x^{3.5} d x

\int x^{1.9} d x = 0.34482 \dots x^{2.9}

\int 9 x^{3.5} d x = 2 x^{4.5}

= 0.34482 \dots x^{2.9} + 2 x^{4.5}

Add a constant to the solution

= 0.34482 \dots x^{2.9} + 2 x^{4.5} + C

\frac{d}{d x} (\frac{x}{x ^{- 1} + 4})

\frac{d}{d x} (1 + 2 e^{x})

f (x) = e^{x^{4}}

\frac{\partial}{\partial x} (\frac{1}{x + y + z})

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

What is the integral of (x^{1.9}+9x^{3.5}) ?
The integral of (x^{1.9}+9x^{3.5}) is 0.34482…x^{2.9}+2x^{4.5}+C