integral of (1+x^4)/(x^3+x)

解答

\int \frac{1 + x ^{4}}{x ^{3} + x} d x

\frac{1}{2} x^{2} - ln x^{2} + 1 + ln ∣ x ∣ + \frac{1}{2} + C

求解步骤

\int \frac{1 + x ^{4}}{x ^{3} + x} d x

乘开

\frac{1 + x ^{4}}{x ^{3} + x} : \frac{1}{x ^{3} + x} + \frac{x ^{4}}{x ^{3} + x}

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{1}{x ^{3} + x} d x + \int \frac{x ^{4}}{x ^{3} + x} d x

\int \frac{1}{x ^{3} + x} d x = ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1

\int \frac{x ^{4}}{x ^{3} + x} d x = \frac{1}{2} (x^{2} + 1 - ln x^{2} + 1)

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 + \frac{1}{2} (x^{2} + 1 - ln x^{2} + 1)

化简

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

= \frac{1}{2} x^{2} - ln x^{2} + 1 + ln ∣ x ∣ + \frac{1}{2}

解答补常数

= \frac{1}{2} x^{2} - ln x^{2} + 1 + ln ∣ x ∣ + \frac{1}{2} + C

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

\int \frac{x ^{3} + 2 x ^{2} + 5 x + 3}{x ^{2} + x + 2} d x

\int \frac{2 x ^{3} - 7 x ^{2} + 3 x - 2}{( x ^{2} - x ) ^{2}} d x

\int \frac{7 x - 9}{x ^{2} - 6 x - 7} d x

\int (\frac{2 x - 1}{x ^{2} - x + 8}) d x