What is the integral of 5x^2+(2/x)^4 ?

The integral of 5x^2+(2/x)^4 is (5x^3)/3-(16)/(3x^3)+C

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integral of 5x^2+(2/x)^4

\int 5 x^{2} + (\frac{2}{x})^{4} d x

\frac{5 x ^{3}}{3} - \frac{16}{3 x ^{3}} + C

Solution steps

\int 5 x^{2} + (\frac{2}{x})^{4} d x

Simplify

5 x^{2} + (\frac{2}{x})^{4} : 5 x^{2} + \frac{16}{x ^{4}}

= \int 5 x^{2} + \frac{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 5 x^{2} d x + \int \frac{16}{x ^{4}} d x

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

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

= \frac{5 x ^{3}}{3} - \frac{16}{3 x ^{3}}

Add a constant to the solution

= \frac{5 x ^{3}}{3} - \frac{16}{3 x ^{3}} + C

\frac{d ^{2} y}{d x ^{2}} + y = 0

y^{^{'}} - \frac{1}{x} y = x^{2} e^{x}

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

\int_{0}^{16} t^{2} + 2 t + 1 d t

y^{^{'}} = \frac{( x ^{3} y ^{3} + 4 x ^{3} )}{y ^{2}}

What is the integral of 5x^2+(2/x)^4 ?
The integral of 5x^2+(2/x)^4 is (5x^3)/3-(16)/(3x^3)+C