What is the integral of ((x^2)/4+4x+19)^2 ?

The integral of ((x^2)/4+4x+19)^2 is (x^5}{80}+\frac{x^4)/2+(17x^3)/2+76x^2+361x+C

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integral of ((x^2)/4+4x+19)^2

\int (\frac{x ^{2}}{4} + 4 x + 19)^{2} d x

\frac{x ^{5}}{80} + \frac{x ^{4}}{2} + \frac{17 x ^{3}}{2} + 76 x^{2} + 361 x + C

Solution steps

\int (\frac{x ^{2}}{4} + 4 x + 19)^{2} d x

Expand

(\frac{x ^{2}}{4} + 4 x + 19)^{2} : \frac{x ^{4}}{16} + 2 x^{3} + \frac{51 x ^{2}}{2} + 152 x + 361

= \int \frac{x ^{4}}{16} + 2 x^{3} + \frac{51 x ^{2}}{2} + 152 x + 361 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{x ^{4}}{16} d x + \int 2 x^{3} d x + \int \frac{51 x ^{2}}{2} d x + \int 152 x d x + \int 361 d x

\int \frac{x ^{4}}{16} d x = \frac{x ^{5}}{80}

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

\int \frac{51 x ^{2}}{2} d x = \frac{17 x ^{3}}{2}

\int 152 x d x = 76 x^{2}

\int 361 d x = 361 x

= \frac{x ^{5}}{80} + \frac{x ^{4}}{2} + \frac{17 x ^{3}}{2} + 76 x^{2} + 361 x

Add a constant to the solution

= \frac{x ^{5}}{80} + \frac{x ^{4}}{2} + \frac{17 x ^{3}}{2} + 76 x^{2} + 361 x + C

(8 y^{2} - x y) d x + x^{2} d y = 0

(e^{x} sin (x))^{^{'}}

\int 13 x^{2} d x

(x^{2} - 5 x + 8) (3 x^{2} - 2)

\frac{1}{s ^{5}}

What is the integral of ((x^2)/4+4x+19)^2 ?
The integral of ((x^2)/4+4x+19)^2 is (x^5}{80}+\frac{x^4)/2+(17x^3)/2+76x^2+361x+C