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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1/(-0.4t^3))

\int (\frac{1}{- 0.4 t ^{3}}) d t

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

Solution steps

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

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}{t ^{3}} d t

Simplify

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

Apply the Power Rule

: - \frac{1}{2 t ^{2}}

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

Simplify

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

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

Add a constant to the solution

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

x \to 2 lim (3 (2 x + 5))

\int (x^{2} + 7 x + 3)^{4} (2 x + 7) d x

h \to 0 lim (\frac{sin ( h )}{6 h})

\frac{d}{d x} (7 + x)

x \to \infty lim (\frac{0}{+ x})

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