What is the integral of 16x^2-x^4 ?

The integral of 16x^2-x^4 is (16x^3)/3-(x^5)/5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 16x^2-x^4

\int 16 x^{2} - x^{4} d x

\frac{16 x ^{3}}{3} - \frac{x ^{5}}{5} + C

Solution steps

\int 16 x^{2} - 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 16 x^{2} d x - \int x^{4} d x

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

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

= \frac{16 x ^{3}}{3} - \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{16 x ^{3}}{3} - \frac{x ^{5}}{5} + C

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

\int \frac{x - 1}{3 x ^{3} + 6 x} d x

\int sec (t) (5 sec (t) + 9 tan (t)) d t

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

\int (\frac{3 x ^{4}}{π} + 10 cos (x) + \frac{1}{e ^{x}}) d x

What is the integral of 16x^2-x^4 ?
The integral of 16x^2-x^4 is (16x^3)/3-(x^5)/5+C