What is area sqrt(x+2), 1/(x+1),0,2 ?

The answer to area sqrt(x+2), 1/(x+1),0,2 is (16-4sqrt(2))/3-ln(3)

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area sqrt(x+2), 1/(x+1),0,2

area

x + 2, \frac{1}{x + 1}, 0, 2

\frac{16 - 4 2}{3} - ln (3)

Decimal Notation

2.34910 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{0}^{2} x + 2 - \frac{1}{x + 1} d x : \frac{16 - 4 2}{3} - ln (3)

The area is:

= \frac{16 - 4 2}{3} - ln (3)

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

x^{2} + 9

\int \frac{42}{1 - cos ( 6 x )} d x

\int (2 x y^{3}) d x

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

What is area sqrt(x+2), 1/(x+1),0,2 ?
The answer to area sqrt(x+2), 1/(x+1),0,2 is (16-4sqrt(2))/3-ln(3)