What is the integral of y/((y^2+1)^{1/2)} ?

The integral of y/((y^2+1)^{1/2)} is (y^2+1)^{1/2}+C

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integral of y/((y^2+1)^{1/2)}

\int \frac{y}{( y ^{2} + 1 ) ^{\frac{1}{2}}} d y

(y^{2} + 1)^{\frac{1}{2}} + C

Solution steps

\int \frac{y}{( y ^{2} + 1 ) ^{\frac{1}{2}}} d y

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int y (y^{2} + 1)^{- \frac{1}{2}} d y

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{1}{2}}} d u

= \int \frac{1}{2 u ^{\frac{1}{2}}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{1}{2}}} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = y^{2} + 1

= \frac{1}{2} \cdot 2 (y^{2} + 1)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (y^{2} + 1)^{\frac{1}{2}} : (y^{2} + 1)^{\frac{1}{2}}

= (y^{2} + 1)^{\frac{1}{2}}

Add a constant to the solution

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

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