What is the integral of 12\sqrt[3]{81-9x}-36 ?

The integral of 12\sqrt[3]{81-9x}-36 is -(81-9x)^{4/3}-36x+C

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integral of 12\sqrt[3]{81-9x}-36

- (81 - 9 x)^{\frac{4}{3}} - 36 x + C

Solution steps

Apply the Sum Rule:

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

\int 36 d x = 36 x

= - (81 - 9 x)^{\frac{4}{3}} - 36 x

Add a constant to the solution

= - (81 - 9 x)^{\frac{4}{3}} - 36 x + C

\int \frac{ln ( t )}{t ^{3}} d t

\int sech^{2} (x) tanh (x) d x

x \to 3 - lim (\frac{2}{x ^{2} - 9})

e^{9 - x}

y^{^{''}} + 25 y = cos (6 t)

What is the integral of 12\sqrt[3]{81-9x}-36 ?
The integral of 12\sqrt[3]{81-9x}-36 is -(81-9x)^{4/3}-36x+C