What is the integral of ((ln(x))^{32})/x ?

The integral of ((ln(x))^{32})/x is (ln^{33}(x))/(33)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ((ln(x))^{32})/x

\int \frac{( ln ( x ) ) ^{32}}{x} d x

\frac{ln ^{33} ( x )}{33} + C

Solution steps

\int \frac{( ln ( x ) ) ^{32}}{x} d x

Apply u-substitution

= \int u^{32} d u

Apply the Power Rule

= \frac{u ^{33}}{33}

Substitute back

u = ln (x)

= \frac{ln ^{33} ( x )}{33}

Add a constant to the solution

= \frac{ln ^{33} ( x )}{33} + C

x \to 2 lim (2 x^{4} - 10)

\frac{d ^{2}}{d x ^{2}} (ln (x^{2} + 3 x + 15))

f (x) = 8 ln (x)

x \to 1 lim (3 x^{2} - 7 x + 1)

\frac{d}{d x} (x^{3} - 6 x^{2} + 8 x)

What is the integral of ((ln(x))^{32})/x ?
The integral of ((ln(x))^{32})/x is (ln^{33}(x))/(33)+C