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

The integral of (x^4-4^x+\sqrt[4]{x}-4/x) is (x^5)/5-(2^{2x-1})/(ln(2))+4/5 x^{5/4}-4ln|x|+C

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

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

\frac{x ^{5}}{5} - \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{4}{5} x^{\frac{5}{4}} - 4 ln ∣ x ∣ + C

Solution steps

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

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

\int 4^{x} d x = \frac{2 ^{2 x - 1}}{ln ( 2 )}

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= \frac{x ^{5}}{5} - \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{4}{5} x^{\frac{5}{4}} - 4 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{2 ^{2 x - 1}}{ln ( 2 )} + \frac{4}{5} x^{\frac{5}{4}} - 4 ln ∣ x ∣ + C

y = \frac{( x + 7 )}{( x - 5 )}

\frac{d}{d x} (\frac{1}{2} \cdot ln (x))

\int r^{2} (\frac{r ^{3}}{9} + 7)^{2} d r

\int \frac{x ^{5}}{( 4 - x ^{6} ) ^{4}} d x

\frac{d}{d x} (\frac{1}{2} (x - 1)^{3} sin (\frac{1}{x - 1}))

What is the integral of (x^4-4^x+\sqrt[4]{x}-4/x) ?
The integral of (x^4-4^x+\sqrt[4]{x}-4/x) is (x^5)/5-(2^{2x-1})/(ln(2))+4/5 x^{5/4}-4ln|x|+C