What is the integral of ((tan(u))/4-4/7 u^6-2/u) ?

The integral of ((tan(u))/4-4/7 u^6-2/u) is -1/4 ln|cos(u)|-(4u^7)/(49)-2ln|u|+C

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integral of ((tan(u))/4-4/7 u^6-2/u)

\int (\frac{tan ( u )}{4} - \frac{4}{7} u^{6} - \frac{2}{u}) d u

- \frac{1}{4} ln ∣ cos (u) ∣ - \frac{4 u ^{7}}{49} - 2 ln ∣ u ∣ + C

Solution steps

\int \frac{tan ( u )}{4} - \frac{4}{7} u^{6} - \frac{2}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{tan ( u )}{4} d u - \int \frac{4}{7} u^{6} d u - \int \frac{2}{u} d u

\int \frac{tan ( u )}{4} d u = - \frac{1}{4} ln ∣ cos (u) ∣

\int \frac{4}{7} u^{6} d u = \frac{4 u ^{7}}{49}

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= - \frac{1}{4} ln ∣ cos (u) ∣ - \frac{4 u ^{7}}{49} - 2 ln ∣ u ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ cos (u) ∣ - \frac{4 u ^{7}}{49} - 2 ln ∣ u ∣ + C

x \to a lim (\frac{1}{2})

f (x) = (5 x + 4)^{3}

\frac{x cos ( x ) - sin ( x )}{7 x ^{2}}

\frac{d}{d x} (x^{3} - 6 x^{2} + 9 x + 4)

\int_{0}^{5} \frac{1}{u ^{2} - 25} d u

What is the integral of ((tan(u))/4-4/7 u^6-2/u) ?
The integral of ((tan(u))/4-4/7 u^6-2/u) is -1/4 ln|cos(u)|-(4u^7)/(49)-2ln|u|+C