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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x^2+1))/(x+1)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} - 2 u + 2}{u} d u

= \int \frac{u ^{2} - 2 u + 2}{u} d u

Expand

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

= \int u - 2 + \frac{2}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u d u - \int 2 d u + \int \frac{2}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 2 d u = 2 u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + 2 ln ∣ u ∣

Substitute back

u = x + 1

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

Add a constant to the solution

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

\int_{0}^{5} \frac{1}{25 + x ^{2}} d x

\int \frac{c}{x ^{2}} d x

f (x) = 7 x^{15}

\int_{0}^{5} \frac{1}{x ^{\frac{18}{19}}} d x

f (x) = \frac{3}{4 x + 1}, at x = - 1

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