What is the integral of (1/4 x^3+3x^2-3/4 x+4) ?

The integral of (1/4 x^3+3x^2-3/4 x+4) is (x^4}{16}+x^3-\frac{3x^2)/8+4x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1/4 x^3+3x^2-3/4 x+4)

\int (\frac{1}{4} x^{3} + 3 x^{2} - \frac{3}{4} x + 4) d x

\frac{x ^{4}}{16} + x^{3} - \frac{3 x ^{2}}{8} + 4 x + C

Solution steps

\int \frac{1}{4} x^{3} + 3 x^{2} - \frac{3}{4} x + 4 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}{4} x^{3} d x + \int 3 x^{2} d x - \int \frac{3}{4} x d x + \int 4 d x

\int \frac{1}{4} x^{3} d x = \frac{x ^{4}}{16}

\int 3 x^{2} d x = x^{3}

\int \frac{3}{4} x d x = \frac{3 x ^{2}}{8}

\int 4 d x = 4 x

= \frac{x ^{4}}{16} + x^{3} - \frac{3 x ^{2}}{8} + 4 x

Add a constant to the solution

= \frac{x ^{4}}{16} + x^{3} - \frac{3 x ^{2}}{8} + 4 x + C

\int e^{8 x} sin (e^{8 x}) d x

\int \frac{7 x}{x ^{4} + 1} d x

\int \frac{e ^{3 x}}{e ^{x} - 1} d x

\frac{1}{ln ( x )}

\int x ln (8 x) d x

What is the integral of (1/4 x^3+3x^2-3/4 x+4) ?
The integral of (1/4 x^3+3x^2-3/4 x+4) is (x^4}{16}+x^3-\frac{3x^2)/8+4x+C