What is the integral of 20t(5t^2-3)^3 ?

The integral of 20t(5t^2-3)^3 is 1/2 (5t^2-3)^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 20t(5t^2-3)^3

\int 20 t (5 t^{2} - 3)^{3} d t

\frac{1}{2} (5 t^{2} - 3)^{4} + C

Solution steps

\int 20 t (5 t^{2} - 3)^{3} d t

Take the constant out:

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

= 20 \cdot \int t (5 t^{2} - 3)^{3} d t

Apply u-substitution

: \int \frac{u ^{3}}{10} d u

= 20 \cdot \int \frac{u ^{3}}{10} d u

Take the constant out:

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

= 20 \cdot \frac{1}{10} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 20 \cdot \frac{1}{10} \cdot \frac{u ^{4}}{4}

Substitute back

u = 5 t^{2} - 3

= 20 \cdot \frac{1}{10} \cdot \frac{( 5 t ^{2} - 3 ) ^{4}}{4}

Simplify

20 \cdot \frac{1}{10} \cdot \frac{( 5 t ^{2} - 3 ) ^{4}}{4} : \frac{1}{2} (5 t^{2} - 3)^{4}

= \frac{1}{2} (5 t^{2} - 3)^{4}

Add a constant to the solution

= \frac{1}{2} (5 t^{2} - 3)^{4} + C

n = 5 \sum \infty \frac{1}{n ln ( n )}

\frac{d}{d x} (\frac{( 2 x ^{3} + 7 ) ( 3 x ^{2} - 5 )}{x ^{2} + 1})

x \to 7 lim (\frac{3}{( x - 7 ) ^{2}})

\frac{\partial}{\partial x} (x^{- 1} y^{2} + x y^{2} + 4 x y)

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

What is the integral of 20t(5t^2-3)^3 ?
The integral of 20t(5t^2-3)^3 is 1/2 (5t^2-3)^4+C