integral of (1+t^2)^{1/2}

解答

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

\frac{1}{2} t (t^{2} + 1)^{\frac{1}{2}} + \frac{1}{2} ln t + 1 + t^{2} + C

求解步骤

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

使用三角换元法

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

使用降次积分法

= \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) ∣

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

u = arctan (t)

代回

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

化简

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

= \frac{1}{2} t (t^{2} + 1)^{\frac{1}{2}} + \frac{1}{2} ln t + 1 + t^{2}

解答补常数

= \frac{1}{2} t (t^{2} + 1)^{\frac{1}{2}} + \frac{1}{2} ln t + 1 + t^{2} + C

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

\int - x^{2} - 2 d x

\int arctan (x - 1) d x

\int x^{2} (8 x^{3} - 9)^{6} d x

\int \frac{x ^{3 + \frac{1}{7}}}{7} d x