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

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

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

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

\frac{3 ln ( 2 ) - ln ( 3 ) + 2 arctan ( \frac{2}{2} )}{6}

Decimal Notation

0.30854 \dots

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{3} + x ^{2} + 2 x + 2} : \frac{1}{3 ( x + 1 )} + \frac{- x + 1}{3 ( x ^{2} + 2 )}

= \int_{0}^{1} \frac{1}{3 ( x + 1 )} + \frac{- x + 1}{3 ( x ^{2} + 2 )} d x

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{1}{3 ( x + 1 )} d x + \int_{0}^{1} \frac{- x + 1}{3 ( x ^{2} + 2 )} d x

\int_{0}^{1} \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln (2)

\int_{0}^{1} \frac{- x + 1}{3 ( x ^{2} + 2 )} d x = \frac{ln ( 2 ) - ln ( 3 ) + 2 arctan ( \frac{1}{2} )}{6}

= \frac{1}{3} ln (2) + \frac{ln ( 2 ) - ln ( 3 ) + 2 arctan ( \frac{1}{2} )}{6}

Simplify

\frac{1}{3} ln (2) + \frac{ln ( 2 ) - ln ( 3 ) + 2 arctan ( \frac{1}{2} )}{6} : \frac{3 ln ( 2 ) - ln ( 3 ) + 2 arctan ( \frac{2}{2} )}{6}

= \frac{3 ln ( 2 ) - ln ( 3 ) + 2 arctan ( \frac{2}{2} )}{6}

\int_{0}^{\frac{π}{2}} 2 π x cos (x) d x

\int_{3}^{6} x^{2} - 3 x d x

\int_{0}^{3} 6 x - x^{2} d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (x) - 1 d x

\int_{0}^{2 π} \frac{1}{2}

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