What is the integral from 0 to 1 of (8+x)/(1+x^2) ?

The integral from 0 to 1 of (8+x)/(1+x^2) is 2pi+1/2 ln(2)

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integral from 0 to 1 of (8+x)/(1+x^2)

\int_{0}^{1} \frac{8 + x}{1 + x ^{2}} d x

2 π + \frac{1}{2} ln (2)

Decimal

6.62975 \dots

Solution steps

\int_{0}^{1} \frac{8 + x}{1 + x ^{2}} d x

Expand

\frac{8 + x}{1 + x ^{2}} : \frac{8}{1 + x ^{2}} + \frac{x}{1 + x ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} \frac{8}{1 + x ^{2}} d x + \int_{0}^{1} \frac{x}{1 + x ^{2}} d x

\int_{0}^{1} \frac{8}{1 + x ^{2}} d x = 2 π

\int_{0}^{1} \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln (2)

= 2 π + \frac{1}{2} ln (2)

\frac{5 + s}{s ( s ^{2} - 5 s + 4 )}

\frac{d}{d x} (\frac{3}{5})

\frac{1 + 2 x}{1 - 2 x}

n = 9 \sum \infty e^{3 - 6 n}

(y^{2} - 1) d x - (2 y + x y) d y = 0

What is the integral from 0 to 1 of (8+x)/(1+x^2) ?
The integral from 0 to 1 of (8+x)/(1+x^2) is 2pi+1/2 ln(2)