integral of t/(sqrt(2t^2+1))

解答

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

\frac{1}{2} 2 t^{2} + 1 + C

求解步骤

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

使用换元积分法

= \int \frac{1}{4 u} d u

提出常数:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

使用幂法则

= \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

u = 2 t^{2} + 1

代回

= \frac{1}{4} \cdot 2 (2 t^{2} + 1)^{\frac{1}{2}}

化简

\frac{1}{4} \cdot 2 (2 t^{2} + 1)^{\frac{1}{2}} : \frac{1}{2} 2 t^{2} + 1

= \frac{1}{2} 2 t^{2} + 1

解答补常数

= \frac{1}{2} 2 t^{2} + 1 + C

\int \frac{9}{x ^{3} + x ^{2}} d x

\int \frac{9 x ^{2} - 10}{( 3 x ^{3} - 10 x ) ^{4}} d x

\int 3 t^{2} t^{3} + 2 d t

\int \frac{π}{x ^{2} + 16} d x

\int \frac{4 x ^{2} + x + 1}{x ^{3} - 1} d x