What is the integral of x^2(2x^3+3)^4 ?

The integral of x^2(2x^3+3)^4 is 1/30 (2x^3+3)^5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2(2x^3+3)^4

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

\frac{1}{30} (2 x^{3} + 3)^{5} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{4}}{6} d u

= \int \frac{u ^{4}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{6} \cdot \frac{u ^{5}}{5}

Substitute back

u = 2 x^{3} + 3

= \frac{1}{6} \cdot \frac{( 2 x ^{3} + 3 ) ^{5}}{5}

Simplify

\frac{1}{6} \cdot \frac{( 2 x ^{3} + 3 ) ^{5}}{5} : \frac{1}{30} (2 x^{3} + 3)^{5}

= \frac{1}{30} (2 x^{3} + 3)^{5}

Add a constant to the solution

= \frac{1}{30} (2 x^{3} + 3)^{5} + C

\int \frac{x + 9}{9 - 3 x - 2 x ^{2}} d x

x \to 0 lim ((1 + \frac{3}{x ^{2}})^{x})

y^{^{''}} + b^{2} y = 0

y^{^{'}} = \frac{( e ^{x - y} )}{1 + e ^{x}}

\int \frac{24}{x ^{2}} d x

What is the integral of x^2(2x^3+3)^4 ?
The integral of x^2(2x^3+3)^4 is 1/30 (2x^3+3)^5+C