What is the integral of 7((ln(z))^2)/z ?

The integral of 7((ln(z))^2)/z is 7/3 ln^3(z)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 7((ln(z))^2)/z

\int 7 \frac{( ln ( z ) ) ^{2}}{z} d z

\frac{7}{3} ln^{3} (z) + C

Solution steps

\int 7 \cdot \frac{( ln ( z ) ) ^{2}}{z} d z

Take the constant out:

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

= 7 \cdot \int \frac{( ln ( z ) ) ^{2}}{z} d z

Apply u-substitution

: \int u^{2} d u

= 7 \cdot \int u^{2} d u

Apply the Power Rule

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

= 7 \cdot \frac{u ^{3}}{3}

Substitute back

u = ln (z)

= 7 \cdot \frac{ln ^{3} ( z )}{3}

Simplify

7 \cdot \frac{ln ^{3} ( z )}{3} : \frac{7}{3} ln^{3} (z)

= \frac{7}{3} ln^{3} (z)

Add a constant to the solution

= \frac{7}{3} ln^{3} (z) + C

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

\int \frac{1}{x ^{7} + 4 x} d x

\int 6 x x d x

\int (4 x - 5)^{\frac{1}{2}} d x

\int 5 x + 6 d x

What is the integral of 7((ln(z))^2)/z ?
The integral of 7((ln(z))^2)/z is 7/3 ln^3(z)+C