What is the integral of 1/(-0.4x^3) ?

The integral of 1/(-0.4x^3) is (1.25)/(x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(-0.4x^3)

\int \frac{1}{- 0.4 x ^{3}} d x

\frac{1.25}{x ^{2}} + C

Solution steps

\int \frac{1}{- 0.4 x ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{- 0.4} \cdot \int \frac{1}{x ^{3}} d x

Simplify

= (- 2.5) \cdot \int \frac{1}{x ^{3}} d x

Apply the Power Rule

= (- 2.5) (- \frac{1}{2 x ^{2}})

Simplify

(- 2.5) (- \frac{1}{2 x ^{2}}) : \frac{1.25}{x ^{2}}

= \frac{1.25}{x ^{2}}

Add a constant to the solution

= \frac{1.25}{x ^{2}} + C

\int sin^{2} (m x) d x

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

\int sec^{3} (t) tan (t) d t

\int (2 e^{x} + 8 x^{7}) d x

\int \frac{x ^{4} + 3 x ^{3} - 2 x ^{2} + 1}{x + 5} d x

What is the integral of 1/(-0.4x^3) ?
The integral of 1/(-0.4x^3) is (1.25)/(x^2)+C