What is the integral of (6ln(x+7))/(x^2) ?

The integral of (6ln(x+7))/(x^2) is 6(-(ln(x+7))/x+1/7 ln|x|-1/7 ln|x+7|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6ln(x+7))/(x^2)

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

6 (- \frac{ln ( x + 7 )}{x} + \frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x + 7 ∣) + C

Solution steps

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

Take the constant out:

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

= 6 \cdot \int \frac{ln ( x + 7 )}{x ^{2}} d x

Apply Integration By Parts

: - \frac{ln ( x + 7 )}{x} - \int - \frac{1}{x ( x + 7 )} d x

= 6 (- \frac{ln ( x + 7 )}{x} - \int - \frac{1}{x ( x + 7 )} d x)

\int - \frac{1}{x ( x + 7 )} d x = - \frac{1}{7} ln ∣ x ∣ + \frac{1}{7} ln ∣ x + 7 ∣

= 6 (- \frac{ln ( x + 7 )}{x} - (- \frac{1}{7} ln ∣ x ∣ + \frac{1}{7} ln ∣ x + 7 ∣))

Simplify

= 6 (- \frac{ln ( x + 7 )}{x} + \frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x + 7 ∣)

Add a constant to the solution

= 6 (- \frac{ln ( x + 7 )}{x} + \frac{1}{7} ln ∣ x ∣ - \frac{1}{7} ln ∣ x + 7 ∣) + C

\frac{\partial}{\partial x} (1 0^{x^{2} + x})

y = \frac{4}{x}

\int_{2}^{8} \frac{1}{x ^{2} - 1} d x

\frac{d}{d x} (\frac{1}{2 x + 3})

x^{2} \cdot cos (x)

What is the integral of (6ln(x+7))/(x^2) ?
The integral of (6ln(x+7))/(x^2) is 6(-(ln(x+7))/x+1/7 ln|x|-1/7 ln|x+7|)+C