What is the integral of x(2x+3)^{99} ?

The integral of x(2x+3)^{99} is 1/4 (((2x+3)^{101})/(101)-(3(2x+3)^{100})/(100))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x(2x+3)^{99}

\int x (2 x + 3)^{99} d x

\frac{1}{4} (\frac{( 2 x + 3 ) ^{101}}{101} - \frac{3 ( 2 x + 3 ) ^{100}}{100}) + C

Solution steps

\int x (2 x + 3)^{99} d x

Apply u-substitution

: \int \frac{u ^{99} ( u - 3 )}{4} d u

= \int \frac{u ^{99} ( u - 3 )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{99} (u - 3) d u

Expand

u^{99} (u - 3) : u^{100} - 3 u^{99}

= \frac{1}{4} \cdot \int u^{100} - 3 u^{99} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int u^{100} d u - \int 3 u^{99} d u)

\int u^{100} d u = \frac{u ^{101}}{101}

\int 3 u^{99} d u = \frac{3 u ^{100}}{100}

= \frac{1}{4} (\frac{u ^{101}}{101} - \frac{3 u ^{100}}{100})

Substitute back

u = 2 x + 3

= \frac{1}{4} (\frac{( 2 x + 3 ) ^{101}}{101} - \frac{3 ( 2 x + 3 ) ^{100}}{100})

Add a constant to the solution

= \frac{1}{4} (\frac{( 2 x + 3 ) ^{101}}{101} - \frac{3 ( 2 x + 3 ) ^{100}}{100}) + C

x^{2} - 7 x

\int_{0}^{8} x 1 + x d x

\frac{1}{6 x - 1}

\int - \frac{6}{π} x d x

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

What is the integral of x(2x+3)^{99} ?
The integral of x(2x+3)^{99} is 1/4 (((2x+3)^{101})/(101)-(3(2x+3)^{100})/(100))+C