What is the integral of 8/(x(ln(x)+1)^{3/2)} ?

The integral of 8/(x(ln(x)+1)^{3/2)} is -(16)/((ln(x)+1)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/(x(ln(x)+1)^{3/2)}

\int \frac{8}{x ( ln ( x ) + 1 ) ^{\frac{3}{2}}} d x

- \frac{16}{( ln ( x ) + 1 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{8}{x ( ln ( x ) + 1 ) ^{\frac{3}{2}}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ( ln ( x ) + 1 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{u ^{\frac{3}{2}}} d u

= 8 \cdot \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

: - \frac{2}{u ^{\frac{1}{2}}}

= 8 (- \frac{2}{u ^{\frac{1}{2}}})

Substitute back

u = ln (x) + 1

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

Simplify

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

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

Add a constant to the solution

= - \frac{16}{( ln ( x ) + 1 ) ^{\frac{1}{2}}} + C

3 x - 4

\int sin (t - s) d s

\frac{5}{4 - x}, at x = - 3

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

\int x^{7} \cdot sin (x^{4}) d x

What is the integral of 8/(x(ln(x)+1)^{3/2)} ?
The integral of 8/(x(ln(x)+1)^{3/2)} is -(16)/((ln(x)+1)^{1/2)}+C