What is the integral of (w^2+z^2)^{(-1)/2}w ?

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

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integral of (w^2+z^2)^{(-1)/2}w

\int (w^{2} + z^{2})^{\frac{- 1}{2}} w d w

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

Solution steps

\int (w^{2} + z^{2})^{\frac{- 1}{2}} w d w

Apply Integration By Parts

: \frac{w ^{2}}{2 ( z ^{2} + w ^{2} ) ^{\frac{1}{2}}} - \int - \frac{w ^{3}}{2 ( z ^{2} + w ^{2} ) ^{\frac{3}{2}}} d w

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

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

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

Simplify

= \frac{w ^{2}}{2 ( z ^{2} + w ^{2} ) ^{\frac{1}{2}}} + \frac{w ^{2} + 2 z ^{2}}{2 ( w ^{2} + z ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

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

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