What is area y=sqrt(x+2),y= 1/(x+1),[0,2] ?

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

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area y=sqrt(x+2),y= 1/(x+1),[0,2]

area

y = x + 2, y = \frac{1}{x + 1}, [0, 2]

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

Decimal

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}{d x} (5 x^{3} + 3 x)

\frac{\partial}{\partial x} (\frac{x ^{3} + 5}{y ^{3} - 1})

\frac{d}{d x} (arccos (e^{3 x}))

\frac{1}{x ^{3}}

\int sec (x) tan (x) - 4 x d x

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