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

The integral of (x^4+2x^2-1)/(sqrt(x)) is 2/9 x^{9/2}+4/5 x^{5/2}-2sqrt(x)+C

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

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

\frac{2}{9} x^{\frac{9}{2}} + \frac{4}{5} x^{\frac{5}{2}} - 2 x + C

Solution steps

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

Expand

\frac{x ^{4} + 2 x ^{2} - 1}{x} : x^{\frac{7}{2}} + 2 x^{\frac{3}{2}} - \frac{1}{x}

= \int x^{\frac{7}{2}} + 2 x^{\frac{3}{2}} - \frac{1}{x} 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^{\frac{7}{2}} d x + \int 2 x^{\frac{3}{2}} d x - \int \frac{1}{x} d x

\int x^{\frac{7}{2}} d x = \frac{2}{9} x^{\frac{9}{2}}

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

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

= \frac{2}{9} x^{\frac{9}{2}} + \frac{4}{5} x^{\frac{5}{2}} - 2 x

Add a constant to the solution

= \frac{2}{9} x^{\frac{9}{2}} + \frac{4}{5} x^{\frac{5}{2}} - 2 x + C

\frac{d}{d x} (\frac{sin ( x ) + cos ( x )}{x ^{3} + x - 2})

n = 21 \sum \infty e^{- 0.06 n}

(x^{2} - 9) \frac{d y}{d x} + x y = 0

\int 16 - 4 x^{2} d x

\frac{1}{2} sec (2 x)

What is the integral of (x^4+2x^2-1)/(sqrt(x)) ?
The integral of (x^4+2x^2-1)/(sqrt(x)) is 2/9 x^{9/2}+4/5 x^{5/2}-2sqrt(x)+C