What is the integral of 3ax*e^{-2bx} ?

The integral of 3ax*e^{-2bx} is a 3/(4b^2)(-2be^{-2bx}x-e^{-2bx})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3ax*e^{-2bx}

\int 3 a x \cdot e^{- 2 b x} d x

a \frac{3}{4 b ^{2}} (- 2 b e^{- 2 b x} x - e^{- 2 b x}) + C

Solution steps

\int 3 a x e^{- 2 b x} d x

Take the constant out:

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

= 3 a \cdot \int x e^{- 2 b x} d x

Apply u-substitution

: \int \frac{ln ( u )}{4 b ^{2}} d u

= 3 a \cdot \int \frac{ln ( u )}{4 b ^{2}} d u

Take the constant out:

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

= 3 a \frac{1}{4 b ^{2}} \cdot \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= 3 a \frac{1}{4 b ^{2}} (u ln (u) - \int 1 d u)

\int 1 d u = u

= 3 a \frac{1}{4 b ^{2}} (u ln (u) - u)

Substitute back

u = e^{- 2 b x}

= 3 a \frac{1}{4 b ^{2}} (e^{- 2 b x} ln (e^{- 2 b x}) - e^{- 2 b x})

Simplify

3 a \frac{1}{4 b ^{2}} (e^{- 2 b x} ln (e^{- 2 b x}) - e^{- 2 b x}) : a \frac{3}{4 b ^{2}} (- 2 b e^{- 2 b x} x - e^{- 2 b x})

= a \frac{3}{4 b ^{2}} (- 2 b e^{- 2 b x} x - e^{- 2 b x})

Add a constant to the solution

= a \frac{3}{4 b ^{2}} (- 2 b e^{- 2 b x} x - e^{- 2 b x}) + C

\int (2 x - 1)^{2} - 9 d x

\int e^{x} 13 + e^{x} d x

\int \frac{- cos ( x )}{sin ^{9} ( x )} d x

\int x cos (- 3 x) d x

\int x^{3} (1 + x^{2})^{\frac{1}{2}} d x

What is the integral of 3ax*e^{-2bx} ?
The integral of 3ax*e^{-2bx} is a 3/(4b^2)(-2be^{-2bx}x-e^{-2bx})+C