What is the integral of (x^3-8x+1)/((2x)^4) ?

The integral of (x^3-8x+1)/((2x)^4) is 1/16 ln|x|+1/(4x^2)-1/(48x^3)+C

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integral of (x^3-8x+1)/((2x)^4)

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

\frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x ^{2}} - \frac{1}{48 x ^{3}} + C

Solution steps

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

Expand

\frac{x ^{3} - 8 x + 1}{( 2 x ) ^{4}} : \frac{1}{16 x} - \frac{1}{2 x ^{3}} + \frac{1}{16 x ^{4}}

= \int \frac{1}{16 x} - \frac{1}{2 x ^{3}} + \frac{1}{16 x ^{4}} d x

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{1}{16 x} d x - \int \frac{1}{2 x ^{3}} d x + \int \frac{1}{16 x ^{4}} d x

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

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

\int \frac{1}{16 x ^{4}} d x = - \frac{1}{48 x ^{3}}

= \frac{1}{16} ln ∣ x ∣ - (- \frac{1}{4 x ^{2}}) - \frac{1}{48 x ^{3}}

Simplify

= \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x ^{2}} - \frac{1}{48 x ^{3}}

Add a constant to the solution

= \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x ^{2}} - \frac{1}{48 x ^{3}} + C

x y^{^{'}} + y = - 2 x y^{2}, y (1) = - 5

\frac{d}{d x} (tanh (x))

\int \frac{1}{3 - 5 x} d x

7 x - x^{2} (1.6)

y = x^{2} - 1, y = 3

What is the integral of (x^3-8x+1)/((2x)^4) ?
The integral of (x^3-8x+1)/((2x)^4) is 1/16 ln|x|+1/(4x^2)-1/(48x^3)+C