integral of (x^2+y^2)^{1/2}

解答

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

\frac{1}{2} x (x^{2} + y^{2})^{\frac{1}{2}} + \frac{y ^{2}}{2} ln \frac{x + ( y ^{2} + x ^{2} ) ^{\frac{1}{2}}}{∣ y ∣} + C

求解步骤

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

使用三角换元法

= \int y^{2} sec^{3} (u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= y^{2} \cdot \int sec^{3} (u) d u

使用降次积分法

= y^{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= y^{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{1}{y} x)

代回

= y^{2} \frac{sec ^{2} ( arctan ( \frac{1}{y} x ) ) sin ( arctan ( \frac{1}{y} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{y} x)) + sec (arctan (\frac{1}{y} x))

化简

y^{2} \frac{sec ^{2} ( arctan ( \frac{1}{y} x ) ) sin ( arctan ( \frac{1}{y} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{y} x)) + sec (arctan (\frac{1}{y} x)) : \frac{1}{2} x (x^{2} + y^{2})^{\frac{1}{2}} + \frac{y ^{2}}{2} ln \frac{x + ( y ^{2} + x ^{2} ) ^{\frac{1}{2}}}{∣ y ∣}

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

解答补常数

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