integral of 6/(x^2+6x)

解答

\int \frac{6}{x ^{2} + 6 x} d x

ln ∣ x ∣ - ln ∣ x + 6 ∣ + C

求解步骤

\int \frac{6}{x ^{2} + 6 x} d x

提出常数:

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

= 6 \cdot \int \frac{1}{x ^{2} + 6 x} d x

将

\frac{1}{x ^{2} + 6 x}

用部份分式展开:

\frac{1}{6 x} - \frac{1}{6 ( x + 6 )}

= 6 \cdot \int \frac{1}{6 x} - \frac{1}{6 ( x + 6 )} d x

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

= 6 (\int \frac{1}{6 x} d x - \int \frac{1}{6 ( x + 6 )} d x)

\int \frac{1}{6 x} d x = \frac{1}{6} ln ∣ x ∣

\int \frac{1}{6 ( x + 6 )} d x = \frac{1}{6} ln ∣ x + 6 ∣

= 6 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{6} ln ∣ x + 6 ∣)

化简

6 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{6} ln ∣ x + 6 ∣) : ln ∣ x ∣ - ln ∣ x + 6 ∣

= ln ∣ x ∣ - ln ∣ x + 6 ∣

解答补常数

= ln ∣ x ∣ - ln ∣ x + 6 ∣ + C

\int \frac{1}{x ( ln ( x ^{3} ) )} d x

n = 1 \sum \infty \frac{5}{n ^{2} + 64}

\int_{1}^{6 x} \frac{7}{t} d t

\int cot^{5} (x) d x

\frac{d}{d x} (\frac{x + 4}{x})