What is the integral of 5xe^{9x} ?

The integral of 5xe^{9x} is 5/81 (9e^{9x}x-e^{9x})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5xe^{9x}

\int 5 x e^{9 x} d x

\frac{5}{81} (9 e^{9 x} x - e^{9 x}) + C

Solution steps

\int 5 x e^{9 x} d x

Take the constant out:

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

= 5 \cdot \int x e^{9 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{81} d u

= 5 \cdot \int \frac{e ^{u} u}{81} d u

Take the constant out:

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

= 5 \cdot \frac{1}{81} \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= 5 \cdot \frac{1}{81} (e^{u} u - \int e^{u} d u)

\int e^{u} d u = e^{u}

= 5 \cdot \frac{1}{81} (e^{u} u - e^{u})

Substitute back

u = 9 x

= 5 \cdot \frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x})

Simplify

5 \cdot \frac{1}{81} (e^{9 x} \cdot 9 x - e^{9 x}) : \frac{5}{81} (9 e^{9 x} x - e^{9 x})

= \frac{5}{81} (9 e^{9 x} x - e^{9 x})

Add a constant to the solution

= \frac{5}{81} (9 e^{9 x} x - e^{9 x}) + C

\int 3 e^{- 5 x} d x

\frac{d}{d x} (4 e^{- 2 x})

\int \frac{cos ( t )}{t} d t

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

\int (x^{2} + 9) e^{- x} d x

What is the integral of 5xe^{9x} ?
The integral of 5xe^{9x} is 5/81 (9e^{9x}x-e^{9x})+C