What is the integral from 0 to 1 of (e^x+1)(e^x-1) ?

The integral from 0 to 1 of (e^x+1)(e^x-1) is (e^2-1)/2-1

Popular >

integral from 0 to 1 of (e^x+1)(e^x-1)

\int_{0}^{1} (e^{x} + 1) (e^{x} - 1) d x

\frac{e ^{2} - 1}{2} - 1

Decimal Notation

2.19452 \dots

Solution steps

\int_{0}^{1} (e^{x} + 1) (e^{x} - 1) d x

Multiply by the conjugate of

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

= \int_{0}^{1} \frac{e ^{2 x} - 1}{e ^{x} - 1} (e^{x} - 1) d x

Simplify

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

= \int_{0}^{1} e^{2 x} - 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_{0}^{1} e^{2 x} d x - \int_{0}^{1} 1 d x

\int_{0}^{1} e^{2 x} d x = \frac{e ^{2} - 1}{2}

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

= \frac{e ^{2} - 1}{2} - 1

\int_{x}^{2 x} e^{x} d x

\int_{0}^{x} \frac{1}{x ^{3}} d x

\int_{1}^{2} \frac{3}{7} x^{4} d x

\int_{0}^{1} x^{8} (1 - x^{3})^{\frac{5}{4}} d x

\int_{- 2}^{2} x 2 - x d x

What is the integral from 0 to 1 of (e^x+1)(e^x-1) ?
The integral from 0 to 1 of (e^x+1)(e^x-1) is (e^2-1)/2-1