What is the integral of (u+4)(2u+1) ?

The integral of (u+4)(2u+1) is (2u^3)/3+(9u^2)/2+4u+C

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integral of (u+4)(2u+1)

\int (u + 4) (2 u + 1) d u

\frac{2 u ^{3}}{3} + \frac{9 u ^{2}}{2} + 4 u + C

Solution steps

\int (u + 4) (2 u + 1) d u

Expand

(u + 4) (2 u + 1) : 2 u^{2} + 9 u + 4

= \int 2 u^{2} + 9 u + 4 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 u^{2} d u + \int 9 u d u + \int 4 d u

\int 2 u^{2} d u = \frac{2 u ^{3}}{3}

\int 9 u d u = \frac{9 u ^{2}}{2}

\int 4 d u = 4 u

= \frac{2 u ^{3}}{3} + \frac{9 u ^{2}}{2} + 4 u

Add a constant to the solution

= \frac{2 u ^{3}}{3} + \frac{9 u ^{2}}{2} + 4 u + C

y^{^{''}} - 2 y^{^{'}} + 2 y = cos (t), y (0) = 1, y^{^{'}} (0) = 0

\int \frac{1}{2 + x ^{2}} d x

\int \frac{1}{x ^{2}} e^{l n (x)} d x

\frac{d}{d x} (\frac{2 x ^{3}}{3} - 8 x - 5)

y = tan (7 x), y = 2 sin (7 x)

What is the integral of (u+4)(2u+1) ?
The integral of (u+4)(2u+1) is (2u^3)/3+(9u^2)/2+4u+C