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

The integral of ((1/x+1))/(x^2) is -((1+x)^2)/(2x^2)+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+1))/(x^2)

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

- \frac{( 1 + x ) ^{2}}{2 x ^{2}} + C

Solution steps

\int \frac{\frac{1}{x} + 1}{x ^{2}} d x

Apply u-substitution

: \int - u d u

= \int - u d u

Take the constant out:

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

= - \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{u ^{2}}{2}

Substitute back

u = \frac{1}{x} + 1

= - \frac{( \frac{1}{x} + 1 ) ^{2}}{2}

Simplify

- \frac{( \frac{1}{x} + 1 ) ^{2}}{2} : - \frac{( 1 + x ) ^{2}}{2 x ^{2}}

= - \frac{( 1 + x ) ^{2}}{2 x ^{2}}

Add a constant to the solution

= - \frac{( 1 + x ) ^{2}}{2 x ^{2}} + C

y = x^{2}, x + y = 2

y^{^{'}} + \frac{1}{2} \cdot y = \frac{3}{2} \cdot t

B (x - 2)^{2} x^{2}

\int \frac{2 x}{x + 5} d x

\int_{0}^{1} 7 x d x

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