What is the integral of 1/(x(ln(x^2))^4) ?

The integral of 1/(x(ln(x^2))^4) is -1/(48ln^3(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/(x(ln(x^2))^4)

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

- \frac{1}{48 ln ^{3} ( x )} + C

Solution steps

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

Simplify

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

= \int \frac{1}{16 x ln ^{4} ( x )} d x

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{x ln ^{4} ( x )} d x

Apply u-substitution

: \int \frac{1}{u ^{4}} d u

= \frac{1}{16} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{16} (- \frac{1}{3 u ^{3}})

Substitute back

u = ln (x)

= \frac{1}{16} (- \frac{1}{3 ln ^{3} ( x )})

Simplify

\frac{1}{16} (- \frac{1}{3 ln ^{3} ( x )}) : - \frac{1}{48 ln ^{3} ( x )}

= - \frac{1}{48 ln ^{3} ( x )}

Add a constant to the solution

= - \frac{1}{48 ln ^{3} ( x )} + C

x \to 0 - lim (\frac{- ln ( 1 + 6 x )}{x})

\int 2 x (x^{2} + 7)^{6} d x

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

\int \frac{1}{x ^{3} + 4 x ^{2} + 8 x} d x

\int \frac{7 - 98 x}{4 - 49 x ^{2}} d x

What is the integral of 1/(x(ln(x^2))^4) ?
The integral of 1/(x(ln(x^2))^4) is -1/(48ln^3(x))+C