What is the integral of (3\sqrt[5]{x^2}+3) ?

The integral of (3\sqrt[5]{x^2}+3) is 15/7 x^{7/5}+3x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3\sqrt[5]{x^2}+3)

\frac{15}{7} x^{\frac{7}{5}} + 3 x + C

Solution steps

Apply the Sum Rule:

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

\int 3 d x = 3 x

= \frac{15}{7} x^{\frac{7}{5}} + 3 x

Add a constant to the solution

= \frac{15}{7} x^{\frac{7}{5}} + 3 x + C

\int \frac{6 x - 4}{x ^{2} - 10 x - 119} d x

\int \frac{z}{e ^{3 z}} d z

\int 12 x 2 x^{2} + 1 d x

\int sin^{3} (2 θ) d θ

\int \frac{x ^{2}}{( 9 + x ^{2} ) ^{\frac{1}{2}}} d x

What is the integral of (3\sqrt[5]{x^2}+3) ?
The integral of (3\sqrt[5]{x^2}+3) is 15/7 x^{7/5}+3x+C