What is the integral of 1/x+x/e-x^e+e^x-2^{e+1} ?

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

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

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

Solution steps

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{x}{e} d x = \frac{x ^{2}}{2 e}

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

\int e^{x} d x = e^{x}

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

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

Add a constant to the solution

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

\int \frac{1}{3 + 5 sin ( x )} d x

\int \frac{cos ^{3} ( x )}{sin ( x ) + cos ( x )} d x

\int - 6 x 3^{x} d x

\int \frac{2 x}{x ^{4} + 2 x ^{2} + 1} d x

\int \frac{1}{( 4 x - 3 ) ^{2}} d x

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