What is the integral of 4z+zsqrt(1-z^2) ?

The integral of 4z+zsqrt(1-z^2) is 2z^2-1/3 (1-z^2)^{3/2}+C

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integral of 4z+zsqrt(1-z^2)

\int 4 z + z 1 - z^{2} d z

2 z^{2} - \frac{1}{3} (1 - z^{2})^{\frac{3}{2}} + C

Solution steps

\int 4 z + z 1 - z^{2} d z

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 z d z + \int z 1 - z^{2} d z

\int 4 z d z = 2 z^{2}

\int z 1 - z^{2} d z = - \frac{1}{3} (1 - z^{2})^{\frac{3}{2}}

= 2 z^{2} - \frac{1}{3} (1 - z^{2})^{\frac{3}{2}}

Add a constant to the solution

= 2 z^{2} - \frac{1}{3} (1 - z^{2})^{\frac{3}{2}} + C

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What is the integral of 4z+zsqrt(1-z^2) ?
The integral of 4z+zsqrt(1-z^2) is 2z^2-1/3 (1-z^2)^{3/2}+C