integral of (4^{2x})/(1+4^{2x)}

解答

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

8 (\frac{8 ln 8 + 2 ^{- 4 x + 3}}{ln ( 2 )} + 2 x) + C

求解步骤

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

使用换元积分法

= \int \frac{8}{1 + 4 ^{u}} d u

提出常数:

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

= 8 \cdot \int \frac{1}{1 + 4 ^{u}} d u

\frac{1}{1 + 4 ^{u}} = - \frac{4 ^{u}}{1 + 4 ^{u}} + 1

= 8 \cdot \int - \frac{4 ^{u}}{1 + 4 ^{u}} + 1 d u

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

= 8 (- \int \frac{4 ^{u}}{1 + 4 ^{u}} d u + \int 1 d u)

\int \frac{4 ^{u}}{1 + 4 ^{u}} d u = - \frac{8 ln 8 + 2 ^{- 2 u + 3}}{ln ( 2 )}

\int 1 d u = u

= 8 (- (- \frac{8 ln 8 + 2 ^{- 2 u + 3}}{ln ( 2 )}) + u)

u = 2 x

代回

= 8 (- (- \frac{8 ln 8 + 2 ^{- 2 \cdot 2 x + 3}}{ln ( 2 )}) + 2 x)

化简

= 8 (\frac{8 ln 8 + 2 ^{- 4 x + 3}}{ln ( 2 )} + 2 x)

解答补常数

= 8 (\frac{8 ln 8 + 2 ^{- 4 x + 3}}{ln ( 2 )} + 2 x) + C

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