What is the integral of 1/2 x2^{-1/4 x} ?

The integral of 1/2 x2^{-1/4 x} is 8/(ln^2(2))(-ln(2)*2^{(-x-8)/4}x-2^{-1/4 x})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/2 x2^{-1/4 x}

\int \frac{1}{2} x 2^{- \frac{1}{4} x} d x

\frac{8}{ln ^{2} ( 2 )} (- ln (2) \cdot 2^{\frac{- x - 8}{4}} x - 2^{- \frac{1}{4} x}) + C

Solution steps

\int \frac{1}{2} x \cdot 2^{- \frac{1}{4} x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x \cdot 2^{- \frac{1}{4} x} d x

Apply u-substitution

: \int \frac{16 ln ( u )}{ln ^{2} ( 2 )} d u

= \frac{1}{2} \cdot \int \frac{16 ln ( u )}{ln ^{2} ( 2 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{16}{ln ^{2} ( 2 )} \cdot \int ln (u) d u

Apply Integration By Parts

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

= \frac{1}{2} \cdot \frac{16}{ln ^{2} ( 2 )} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} \cdot \frac{16}{ln ^{2} ( 2 )} (u ln (u) - u)

Substitute back

u = 2^{- \frac{1}{4} x}

= \frac{1}{2} \cdot \frac{16}{ln ^{2} ( 2 )} (2^{- \frac{1}{4} x} ln (2^{- \frac{1}{4} x}) - 2^{- \frac{1}{4} x})

Simplify

\frac{1}{2} \cdot \frac{16}{ln ^{2} ( 2 )} (2^{- \frac{1}{4} x} ln (2^{- \frac{1}{4} x}) - 2^{- \frac{1}{4} x}) : \frac{8}{ln ^{2} ( 2 )} (- ln (2) \cdot 2^{\frac{- x - 8}{4}} x - 2^{- \frac{1}{4} x})

= \frac{8}{ln ^{2} ( 2 )} (- ln (2) \cdot 2^{\frac{- x - 8}{4}} x - 2^{- \frac{1}{4} x})

Add a constant to the solution

= \frac{8}{ln ^{2} ( 2 )} (- ln (2) \cdot 2^{\frac{- x - 8}{4}} x - 2^{- \frac{1}{4} x}) + C

\int 4 cos (9 x) d x

\int x (3 x + 5)^{11} d x

\int \frac{1}{x ^{2} + 10 x + 61} d x

\int sec^{6} (θ) d θ

\int (x - 2) (3 - x) d x

What is the integral of 1/2 x2^{-1/4 x} ?
The integral of 1/2 x2^{-1/4 x} is 8/(ln^2(2))(-ln(2)*2^{(-x-8)/4}x-2^{-1/4 x})+C