What is the integral of (7/(x^5)+4x+3) ?

The integral of (7/(x^5)+4x+3) is -7/(4x^4)+2x^2+3x+C

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integral of (7/(x^5)+4x+3)

\int (\frac{7}{x ^{5}} + 4 x + 3) d x

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

Solution steps

\int \frac{7}{x ^{5}} + 4 x + 3 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{7}{x ^{5}} d x + \int 4 x d x + \int 3 d x

\int \frac{7}{x ^{5}} d x = - \frac{7}{4 x ^{4}}

\int 4 x d x = 2 x^{2}

\int 3 d x = 3 x

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

Add a constant to the solution

= - \frac{7}{4 x ^{4}} + 2 x^{2} + 3 x + C

ln (x x)

18 x - 3 x

\frac{d y}{d t} + 4 y = 2 e^{- 3 t}

\int e^{6 x} cos (x) d x

x \to - 3 lim (2 x + 6)

What is the integral of (7/(x^5)+4x+3) ?
The integral of (7/(x^5)+4x+3) is -7/(4x^4)+2x^2+3x+C