What is the integral of e^{ln((2x^2+2x+1)/(x+1))} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{ln((2x^2+2x+1)/(x+1))}

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

x^{2} + ln ∣ x + 1 ∣ - 1 + C

Solution steps

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

Apply log rule:

a^{l o g_{a} (b)} = b

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

Expand

\frac{2 x ^{2} + 2 x + 1}{x + 1} : \frac{2 x ^{2}}{x + 1} + \frac{2 x}{x + 1} + \frac{1}{x + 1}

Apply the Sum Rule:

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

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

\int \frac{2 x ^{2}}{x + 1} d x = x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3

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

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

= x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3 + 2 (x + 1 - ln ∣ x + 1 ∣) + ln ∣ x + 1 ∣

Simplify

x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3 + 2 (x + 1 - ln ∣ x + 1 ∣) + ln ∣ x + 1 ∣ : x^{2} + ln ∣ x + 1 ∣ - 1

= x^{2} + ln ∣ x + 1 ∣ - 1

Add a constant to the solution

= x^{2} + ln ∣ x + 1 ∣ - 1 + C

\int \frac{1}{x ( 200 - x )} d x

\int (\frac{1 - 2 x}{3})^{3} d x

\int \frac{5 x ^{2} - 12}{x ^{2} - 6 x + 13} d x

\int - \frac{2 z}{3} d z

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

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