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受欢迎的微积分 >

integral of 7/((t-1)sqrt(t^2-2t-48))

解答

\int \frac{7}{( t - 1 ) t ^{2} - 2 t - 48} d t

解答

arcsec (\frac{1}{7} (t - 1)) + C

求解步骤

\int \frac{7}{( t - 1 ) t ^{2} - 2 t - 48} d t

提出常数:

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

= 7 \cdot \int \frac{1}{( t - 1 ) t ^{2} - 2 t - 48} d t

对

t^{2} - 2 t - 48

配方:

(t - 1)^{2} - 49

= 7 \cdot \int \frac{1}{( t - 1 ) ( t - 1 ) ^{2} - 49} d t

使用换元积分法

= 7 \cdot \int \frac{1}{u u ^{2} - 49} d u

使用三角换元法

= 7 \cdot \int \frac{1}{7} d v

提出常数:

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

= 7 \cdot \frac{1}{7} \cdot \int 1 d v

常数积分:

\int a d x = a x

= 7 \cdot \frac{1}{7} \cdot 1 \cdot v

代回

= 7 \cdot \frac{1}{7} \cdot 1 \cdot arcsec (\frac{1}{7} (t - 1))

化简

7 \cdot \frac{1}{7} \cdot 1 \cdot arcsec (\frac{1}{7} (t - 1)) : arcsec (\frac{1}{7} (t - 1))

= arcsec (\frac{1}{7} (t - 1))

解答补常数

= arcsec (\frac{1}{7} (t - 1)) + C

作图

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