integral of (y^2+6y+4)/(y^4+5y^2+4)

解答

\int \frac{y ^{2} + 6 y + 4}{y ^{4} + 5 y ^{2} + 4} d y

arctan (y) + ln y^{2} + 1 - ln y^{2} + 4 + C

求解步骤

\int \frac{y ^{2} + 6 y + 4}{y ^{4} + 5 y ^{2} + 4} d y

乘开

\frac{y ^{2} + 6 y + 4}{y ^{4} + 5 y ^{2} + 4} : \frac{y ^{2}}{y ^{4} + 5 y ^{2} + 4} + \frac{6 y}{y ^{4} + 5 y ^{2} + 4} + \frac{4}{y ^{4} + 5 y ^{2} + 4}

使用积分加法定则:

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

= \int \frac{y ^{2}}{y ^{4} + 5 y ^{2} + 4} d y + \int \frac{6 y}{y ^{4} + 5 y ^{2} + 4} d y + \int \frac{4}{y ^{4} + 5 y ^{2} + 4} d y

\int \frac{y ^{2}}{y ^{4} + 5 y ^{2} + 4} d y = - \frac{1}{3} arctan (y) + \frac{2}{3} arctan (\frac{y}{2})

\int \frac{6 y}{y ^{4} + 5 y ^{2} + 4} d y = ln y^{2} + 1 - ln y^{2} + 4

\int \frac{4}{y ^{4} + 5 y ^{2} + 4} d y = 4 (\frac{1}{3} arctan (y) - \frac{1}{6} arctan (\frac{y}{2}))

= - \frac{1}{3} arctan (y) + \frac{2}{3} arctan (\frac{y}{2}) + ln y^{2} + 1 - ln y^{2} + 4 + 4 (\frac{1}{3} arctan (y) - \frac{1}{6} arctan (\frac{y}{2}))

化简

- \frac{1}{3} arctan (y) + \frac{2}{3} arctan (\frac{y}{2}) + ln y^{2} + 1 - ln y^{2} + 4 + 4 (\frac{1}{3} arctan (y) - \frac{1}{6} arctan (\frac{y}{2})) : arctan (y) + ln y^{2} + 1 - ln y^{2} + 4

= arctan (y) + ln y^{2} + 1 - ln y^{2} + 4

解答补常数

= arctan (y) + ln y^{2} + 1 - ln y^{2} + 4 + C

\int x^{4} - 13 d x

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