What is the integral of (-0.08)(x-6.5)^3+0.3 ?

The integral of (-0.08)(x-6.5)^3+0.3 is -0.02(x-6.5)^4+0.3x+C

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integral of (-0.08)(x-6.5)^3+0.3

\int (- 0.08) (x - 6.5)^{3} + 0.3 d x

- 0.02 (x - 6.5)^{4} + 0.3 x + C

Solution steps

\int (- 0.08) (x - 6.5)^{3} + 0.3 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 (- 0.08) (x - 6.5)^{3} d x + \int 0.3 d x

\int (- 0.08) (x - 6.5)^{3} d x = - 0.02 (x - 6.5)^{4}

\int 0.3 d x = 0.3 x

= - 0.02 (x - 6.5)^{4} + 0.3 x

Add a constant to the solution

= - 0.02 (x - 6.5)^{4} + 0.3 x + C

\int \frac{ln ( x )}{( x )} d x

\int \frac{3 x + 4}{x ( x + 2 ) ^{2}} d x

\int \frac{1}{( a ^{2} + y ^{2} ) ^{\frac{3}{2}}} d y

\int \frac{t ^{4} + 4}{t ^{4}} d t

What is the integral of (-0.08)(x-6.5)^3+0.3 ?
The integral of (-0.08)(x-6.5)^3+0.3 is -0.02(x-6.5)^4+0.3x+C