What is the integral of (x^4+5x^3-2x+1)/(sqrt(x^4)) ?

The integral of (x^4+5x^3-2x+1)/(sqrt(x^4)) is (x^3)/3+(5x^2)/2-2ln|x|-1/x+C

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integral of (x^4+5x^3-2x+1)/(sqrt(x^4))

\int \frac{x ^{4} + 5 x ^{3} - 2 x + 1}{x ^{4}} d x

\frac{x ^{3}}{3} + \frac{5 x ^{2}}{2} - 2 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

\int \frac{x ^{4} + 5 x ^{3} - 2 x + 1}{x ^{4}} d x

x^{4} = x^{2},

assuming

x \geq 0

= \int \frac{x ^{4} + 5 x ^{3} - 2 x + 1}{x ^{2}} d x

Expand

\frac{x ^{4} + 5 x ^{3} - 2 x + 1}{x ^{2}} : x^{2} + 5 x - \frac{2}{x} + \frac{1}{x ^{2}}

= \int x^{2} + 5 x - \frac{2}{x} + \frac{1}{x ^{2}} 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 x^{2} d x + \int 5 x d x - \int \frac{2}{x} d x + \int \frac{1}{x ^{2}} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 5 x d x = \frac{5 x ^{2}}{2}

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= \frac{x ^{3}}{3} + \frac{5 x ^{2}}{2} - 2 ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{5 x ^{2}}{2} - 2 ln ∣ x ∣ - \frac{1}{x} + C

\int 200 e^{0.1 x} + 120 e^{- 0.3 x} d x

\int \frac{sin ( 2 x )}{2 + cos ^{2} ( 2 x )} d x

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

\int (\frac{1}{2 x - 1})^{2} d x

\int \frac{1}{2} 1 - 5 x^{2} d x

What is the integral of (x^4+5x^3-2x+1)/(sqrt(x^4)) ?
The integral of (x^4+5x^3-2x+1)/(sqrt(x^4)) is (x^3)/3+(5x^2)/2-2ln|x|-1/x+C