What is the integral of (2x^2+5)(x^2-3x) ?

The integral of (2x^2+5)(x^2-3x) is (2x^5)/5-(3x^4)/2+(5x^3)/3-(15x^2)/2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^2+5)(x^2-3x)

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

\frac{2 x ^{5}}{5} - \frac{3 x ^{4}}{2} + \frac{5 x ^{3}}{3} - \frac{15 x ^{2}}{2} + C

Solution steps

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

Expand

(2 x^{2} + 5) (x^{2} - 3 x) : 2 x^{4} - 6 x^{3} + 5 x^{2} - 15 x

= \int 2 x^{4} - 6 x^{3} + 5 x^{2} - 15 x 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 2 x^{4} d x - \int 6 x^{3} d x + \int 5 x^{2} d x - \int 15 x d x

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

\int 15 x d x = \frac{15 x ^{2}}{2}

= \frac{2 x ^{5}}{5} - \frac{3 x ^{4}}{2} + \frac{5 x ^{3}}{3} - \frac{15 x ^{2}}{2}

Add a constant to the solution

= \frac{2 x ^{5}}{5} - \frac{3 x ^{4}}{2} + \frac{5 x ^{3}}{3} - \frac{15 x ^{2}}{2} + C

\int \frac{1}{y ^{3}} d y

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} + 2 x - 6}{( x + 1 ) ^{2}})

\int_{0}^{\frac{π}{24}} tan (8 x) d x

e^{1 + x^{2}} + 3 x^{4}

\frac{d y}{d x} y = sin (- 1 (6 x + 1))

What is the integral of (2x^2+5)(x^2-3x) ?
The integral of (2x^2+5)(x^2-3x) is (2x^5)/5-(3x^4)/2+(5x^3)/3-(15x^2)/2+C