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

The integral of (x^2+3x-4) is (x^3)/3+(3x^2)/2-4x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

\int (2 x^{3} - \frac{x ^{2}}{2} + 5 x) d x

\int \frac{1}{( sin ( x ) cos ( x ) ) ^{2}} d x

\int \frac{1}{( x ) ( x - 1 )} d x

\int x \cdot arctan (7 x) d x

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

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