What is the integral from 0 to 1 of (6+x)/(1+x^2) ?

The integral from 0 to 1 of (6+x)/(1+x^2) is (3pi+ln(2))/2

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integral from 0 to 1 of (6+x)/(1+x^2)

\int_{0}^{1} \frac{6 + x}{1 + x ^{2}} d x

\frac{3 π + ln ( 2 )}{2}

Decimal Notation

5.05896 \dots

Solution steps

\int_{0}^{1} \frac{6 + x}{1 + x ^{2}} d x

Expand

\frac{6 + x}{1 + x ^{2}} : \frac{6}{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{6}{1 + x ^{2}} d x + \int_{0}^{1} \frac{x}{1 + x ^{2}} d x

\int_{0}^{1} \frac{6}{1 + x ^{2}} d x = \frac{3 π}{2}

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

= \frac{3 π}{2} + \frac{1}{2} ln (2)

Simplify

\frac{3 π}{2} + \frac{1}{2} ln (2) : \frac{3 π + ln ( 2 )}{2}

= \frac{3 π + ln ( 2 )}{2}

f (x) = \frac{x ^{3}}{x + 1}, at x = 1

\frac{d}{d x} (tan (π x - 1) + e^{c o s (x)})

y = (x - 1)^{3}

\int tan (x) ln (cos (x)) d x

\int \frac{x ^{2}}{x ^{2} + 4} d x

What is the integral from 0 to 1 of (6+x)/(1+x^2) ?
The integral from 0 to 1 of (6+x)/(1+x^2) is (3pi+ln(2))/2