What is the integral of x^{1/8}+y^{1/8} ?

The integral of x^{1/8}+y^{1/8} is y^{1/8}x+8/9 x^{9/8}+C

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integral of x^{1/8}+y^{1/8}

\int x^{\frac{1}{8}} + y^{\frac{1}{8}}

y^{\frac{1}{8}} x + \frac{8}{9} x^{\frac{9}{8}} + C

Solution steps

\int x^{\frac{1}{8}} + y^{\frac{1}{8}} 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 y^{\frac{1}{8}} d x + \int x^{\frac{1}{8}} d x

\int y^{\frac{1}{8}} d x = y^{\frac{1}{8}} x

\int x^{\frac{1}{8}} d x = \frac{8}{9} x^{\frac{9}{8}}

= y^{\frac{1}{8}} x + \frac{8}{9} x^{\frac{9}{8}}

Add a constant to the solution

= y^{\frac{1}{8}} x + \frac{8}{9} x^{\frac{9}{8}} + C

y^{^{''}} + k \cdot y = 0

\frac{\partial}{\partial x} (2 e^{4 x y})

\frac{\partial}{\partial x} (ln ((x^{2} + y^{2})^{2}))

\int 9 z 3 z^{2} - 7 d z

\frac{30}{( 1 - x ) ^{4}}

What is the integral of x^{1/8}+y^{1/8} ?
The integral of x^{1/8}+y^{1/8} is y^{1/8}x+8/9 x^{9/8}+C