integral of ((ln(x))^6+1)/x

解答

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

\frac{ln ^{7} ( x )}{7} + ln ∣ x ∣ + C

求解步骤

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

乘开

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

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

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

\int \frac{ln ^{6} ( x )}{x} d x = \frac{ln ^{7} ( x )}{7}

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{ln ^{7} ( x )}{7} + ln ∣ x ∣

解答补常数

= \frac{ln ^{7} ( x )}{7} + ln ∣ x ∣ + C

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