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

The integral from 0 to 1 of 1/(3x^2-6x+6) is pi/(12)

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integral from 0 to 1 of 1/(3x^2-6x+6)

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

\frac{π}{12}

Decimal Notation

0.26179 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{x}{3 x ^{2} - 6 x + 6} - \int - \frac{2 x ( x - 1 )}{3 ( x ^{2} - 2 x + 2 ) ^{2}} d x]_{0}^{1}

= [\frac{x}{3 x ^{2} - 6 x + 6} - \int - \frac{2 x ( x - 1 )}{3 ( x ^{2} - 2 x + 2 ) ^{2}} d x]_{0}^{1}

\int - \frac{2 x ( x - 1 )}{3 ( x ^{2} - 2 x + 2 ) ^{2}} d x = - \frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{3 ( x ^{2} - 2 x + 2 )} + \frac{1}{3} sin (2 arctan (x - 1)) + \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{3 ( x ^{2} - 2 x + 2 )}

= [\frac{x}{3 x ^{2} - 6 x + 6} - (- \frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{3 ( x ^{2} - 2 x + 2 )} + \frac{1}{3} sin (2 arctan (x - 1)) + \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{3 ( x ^{2} - 2 x + 2 )})]_{0}^{1}

Simplify

[\frac{x}{3 x ^{2} - 6 x + 6} - (- \frac{x ( x + arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) - 1 )}{3 ( x ^{2} - 2 x + 2 )} + \frac{1}{3} sin (2 arctan (x - 1)) + \frac{( x - 1 ) ( x ^{2} - 2 x + 2 ) ( \frac{x - 1}{x ^{2} - 2 x + 2} + arctan ( x - 1 ) ) + 1}{3 ( x ^{2} - 2 x + 2 )})]_{0}^{1} : [\frac{x ^{4} arctan ( x - 1 ) - x ^{4} sin ( 2 arctan ( x - 1 ) ) + 2 x ^{3} + 4 x ^{3} sin ( 2 arctan ( x - 1 ) ) - 4 x ^{3} arctan ( x - 1 ) + 8 x ^{2} arctan ( x - 1 ) - 6 x ^{2} - 8 x ^{2} sin ( 2 arctan ( x - 1 ) ) + 8 x sin ( 2 arctan ( x - 1 ) ) + 8 x - 8 x arctan ( x - 1 ) + 4 arctan ( x - 1 ) - 4 sin ( 2 arctan ( x - 1 ) ) - 4}{3 ( x ^{2} - 2 x + 2 ) ^{2}}]_{0}^{1}

= [\frac{x ^{4} arctan ( x - 1 ) - x ^{4} sin ( 2 arctan ( x - 1 ) ) + 2 x ^{3} + 4 x ^{3} sin ( 2 arctan ( x - 1 ) ) - 4 x ^{3} arctan ( x - 1 ) + 8 x ^{2} arctan ( x - 1 ) - 6 x ^{2} - 8 x ^{2} sin ( 2 arctan ( x - 1 ) ) + 8 x sin ( 2 arctan ( x - 1 ) ) + 8 x - 8 x arctan ( x - 1 ) + 4 arctan ( x - 1 ) - 4 sin ( 2 arctan ( x - 1 ) ) - 4}{3 ( x ^{2} - 2 x + 2 ) ^{2}}]_{0}^{1}

Compute the boundaries:

\frac{π}{12}

= \frac{π}{12}

\int_{1}^{2} \frac{2}{( 2 x - 5 ) ^{3}} d x

\int_{2}^{4} \frac{1}{x ^{2}} ln (x) d x

\int_{0}^{1} 2 x e^{2 x} d x

\int_{1}^{5} \frac{1}{x} d x

\int_{0}^{2 π} 6 t 2 d t

What is the integral from 0 to 1 of 1/(3x^2-6x+6) ?
The integral from 0 to 1 of 1/(3x^2-6x+6) is pi/(12)