integral from 0 to 4 of x/(1+x)

解答

\int_{0}^{4} \frac{x}{1 + x} d x

4 - ln (5)

十进制

2.39056 \dots

求解步骤

\int_{0}^{4} \frac{x}{1 + x} d x

使用换元积分法

= \int_{1}^{5} \frac{u - 1}{u} d u

乘开

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \int_{1}^{5} 1 - \frac{1}{u} 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

= \int_{1}^{5} 1 d u - \int_{1}^{5} \frac{1}{u} d u

\int_{1}^{5} 1 d u = 4

\int_{1}^{5} \frac{1}{u} d u = ln (5)

= 4 - ln (5)

\int_{1}^{7} \frac{9}{1 + x ^{2}} d x

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

\int_{0}^{t} e^{4 t} d t

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\int_{- 1}^{1} 3 x^{3} - 2 x + 5 d x