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

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

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

area

y = x + 2, y = \frac{1}{x + 1}, x = 0, x = 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 t} = y^{1.01}

\int \frac{ln ( x )}{x ^{1.2}} d x

x \to - 1 lim (2 x + 2)

\int sin (x) \cdot ln (cos (x)) d x

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