What is the integral of (3\sqrt[4]{x^3}-8x^5+6e^x-2) ?

The integral of (3\sqrt[4]{x^3}-8x^5+6e^x-2) is 12/7 x^{7/4}-(4x^6)/3+6e^x-2x+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[4]{x^3}-8x^5+6e^x-2)

\frac{12}{7} x^{\frac{7}{4}} - \frac{4 x ^{6}}{3} + 6 e^{x} - 2 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 8 x^{5} d x = \frac{4 x ^{6}}{3}

\int 6 e^{x} d x = 6 e^{x}

\int 2 d x = 2 x

= \frac{12}{7} x^{\frac{7}{4}} - \frac{4 x ^{6}}{3} + 6 e^{x} - 2 x

Add a constant to the solution

= \frac{12}{7} x^{\frac{7}{4}} - \frac{4 x ^{6}}{3} + 6 e^{x} - 2 x + C

\frac{d y}{d x} = y (3 - y)

x \to + (- 1) - lim (x^{2} + 1)

\frac{d}{d x} (+ 2 x)

f (x) = 6 x, (9, 18)

f (x) = tan^{3} (x)

What is the integral of (3\sqrt[4]{x^3}-8x^5+6e^x-2) ?
The integral of (3\sqrt[4]{x^3}-8x^5+6e^x-2) is 12/7 x^{7/4}-(4x^6)/3+6e^x-2x+C