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

4.36581 \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}

x \to 2 lim (x + 2)

\int \frac{1}{( 361 + x ^{2} ) ^{\frac{3}{2}}} d x

\int \frac{( x ^{2} + 1 )}{x ^{2} ( x + 1 ) ^{3}} d x

\int tan (11 x se) c^{2} 11 x 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