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

The integral from 0 to 1 of pi(e^x+1)^2 is 2pie+pi(e^2-3)/2

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integral from 0 to 1 of pi(e^x+1)^2

\int_{0}^{1} π (e^{x} + 1)^{2} d x

2 π e + π \frac{e ^{2} - 3}{2}

Decimal

23.97378 \dots

Solution steps

\int_{0}^{1} π (e^{x} + 1)^{2} d x

Take the constant out:

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

= π \cdot \int_{0}^{1} (e^{x} + 1)^{2} d x

Apply u-substitution

: \int_{1}^{e} \frac{( u + 1 ) ^{2}}{u} d u

= π \cdot \int_{1}^{e} \frac{( u + 1 ) ^{2}}{u} d u

Expand

\frac{( u + 1 ) ^{2}}{u} : u + 2 + \frac{1}{u}

= π \cdot \int_{1}^{e} u + 2 + \frac{1}{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

= π (\int_{1}^{e} u d u + \int_{1}^{e} 2 d u + \int_{1}^{e} \frac{1}{u} d u)

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

\int_{1}^{e} 2 d u = 2 e - 2

\int_{1}^{e} \frac{1}{u} d u = 1

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

Simplify

π (\frac{e ^{2} - 1}{2} + 2 e - 2 + 1) : 2 π e + π \frac{e ^{2} - 3}{2}

= 2 π e + π \frac{e ^{2} - 3}{2}

2 y \frac{d y}{d x} + y^{2} cot (x) = csc (x)

x \to 0 lim ((2 x - 2)^{3})

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

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

(3 x - 2)^{6}

What is the integral from 0 to 1 of pi(e^x+1)^2 ?
The integral from 0 to 1 of pi(e^x+1)^2 is 2pie+pi(e^2-3)/2