What is the integral of (x^4+(16)/(\sqrt[4]{x)})^2 ?

The integral of (x^4+(16)/(\sqrt[4]{x)})^2 is 512sqrt(x)+128/19 x^{19/4}+(x^9)/9+C

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integral of (x^4+(16)/(\sqrt[4]{x)})^2

\int (x^{4} + \frac{16}{4 x})^{2} d x

512 x + \frac{128}{19} x^{\frac{19}{4}} + \frac{x ^{9}}{9} + C

Solution steps

\int (x^{4} + \frac{16}{4 x})^{2} d x

Expand

(x^{4} + \frac{16}{4 x})^{2} : \frac{256}{x} + 32 x^{\frac{15}{4}} + x^{8}

= \int \frac{256}{x} + 32 x^{\frac{15}{4}} + x^{8} 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{256}{x} d x + \int 32 x^{\frac{15}{4}} d x + \int x^{8} d x

\int \frac{256}{x} d x = 512 x

\int 32 x^{\frac{15}{4}} d x = \frac{128}{19} x^{\frac{19}{4}}

\int x^{8} d x = \frac{x ^{9}}{9}

= 512 x + \frac{128}{19} x^{\frac{19}{4}} + \frac{x ^{9}}{9}

Add a constant to the solution

= 512 x + \frac{128}{19} x^{\frac{19}{4}} + \frac{x ^{9}}{9} + C

- x^{2} + 2 x + a

x \to c lim (15)

x \to 0 lim (\frac{sin ( x )}{1})

\int \frac{( x ^{2} + 7 x + 1 )}{( x ^{2} + 1 ) ^{2}} d x

{t^{2} - 3 t - 2 e^{- t} sin (3 t)}

What is the integral of (x^4+(16)/(\sqrt[4]{x)})^2 ?
The integral of (x^4+(16)/(\sqrt[4]{x)})^2 is 512sqrt(x)+128/19 x^{19/4}+(x^9)/9+C