What is the integral of 1/(9x(ln(7x))^5) ?

The integral of 1/(9x(ln(7x))^5) is -1/(36ln^4(7x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(9x(ln(7x))^5)

\int \frac{1}{9 x ( ln ( 7 x ) ) ^{5}} d x

- \frac{1}{36 ln ^{4} ( 7 x )} + C

Solution steps

\int \frac{1}{9 x ( ln ( 7 x ) ) ^{5}} d x

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{x ( ln ( 7 x ) ) ^{5}} d x

Simplify

= \frac{1}{9} \cdot \int \frac{1}{x ln ^{5} ( 7 x )} d x

Apply u-substitution

: \int \frac{1}{u ln ^{5} ( u )} d u

= \frac{1}{9} \cdot \int \frac{1}{u ln ^{5} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{5}} d v

= \frac{1}{9} \cdot \int \frac{1}{v ^{5}} d v

Apply the Power Rule

: - \frac{1}{4 v ^{4}}

= \frac{1}{9} (- \frac{1}{4 v ^{4}})

Substitute back

= \frac{1}{9} (- \frac{1}{4 ln ^{4} ( 7 x )})

Simplify

\frac{1}{9} (- \frac{1}{4 ln ^{4} ( 7 x )}) : - \frac{1}{36 ln ^{4} ( 7 x )}

= - \frac{1}{36 ln ^{4} ( 7 x )}

Add a constant to the solution

= - \frac{1}{36 ln ^{4} ( 7 x )} + C

x y^{^{'}} + 2 y = e^{x^{2}}

\frac{d}{d x} (- 2 cot (x) csc (x))

2 x^{2} - 5

h (x) = (x - 2) (2 x + 3)

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

What is the integral of 1/(9x(ln(7x))^5) ?
The integral of 1/(9x(ln(7x))^5) is -1/(36ln^4(7x))+C