What is the integral of (4x^3+(81)/(8x^4)) ?

The integral of (4x^3+(81)/(8x^4)) is x^4-(27)/(8x^3)+C

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integral of (4x^3+(81)/(8x^4))

\int (4 x^{3} + \frac{81}{8 x ^{4}}) d x

x^{4} - \frac{27}{8 x ^{3}} + C

Solution steps

\int 4 x^{3} + \frac{81}{8 x ^{4}} 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 4 x^{3} d x + \int \frac{81}{8 x ^{4}} d x

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

\int \frac{81}{8 x ^{4}} d x = - \frac{27}{8 x ^{3}}

= x^{4} - \frac{27}{8 x ^{3}}

Add a constant to the solution

= x^{4} - \frac{27}{8 x ^{3}} + C

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What is the integral of (4x^3+(81)/(8x^4)) ?
The integral of (4x^3+(81)/(8x^4)) is x^4-(27)/(8x^3)+C