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

The integral from 0 to 2 of (x^2+2)/(x+1) is 3ln(3)

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

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

3 ln (3)

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{u ^{2} - 2 u + 3}{u} d u

= \int_{1}^{3} \frac{u ^{2} - 2 u + 3}{u} d u

Expand

\frac{u ^{2} - 2 u + 3}{u} : u - 2 + \frac{3}{u}

= \int_{1}^{3} u - 2 + \frac{3}{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

= \int_{1}^{3} u d u - \int_{1}^{3} 2 d u + \int_{1}^{3} \frac{3}{u} d u

\int_{1}^{3} u d u = 4

\int_{1}^{3} 2 d u = 4

\int_{1}^{3} \frac{3}{u} d u = 3 ln (3)

= 4 - 4 + 3 ln (3)

Simplify

= 3 ln (3)

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

\int_{0}^{4} x e^{x + 2} d x

\int \frac{5 - 3 sin ( 2 x )}{sec ( 6 x )} d x

\int (\frac{x ^{2} + 1}{x}) d x

\frac{1}{4 + x ^{2}}

What is the integral from 0 to 2 of (x^2+2)/(x+1) ?
The integral from 0 to 2 of (x^2+2)/(x+1) is 3ln(3)