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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int (x sin (x^{2})) d x

\int_{1}^{4} \frac{3}{t ^{4}} d t

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = 0

\frac{d}{d x} (8 x^{5} 2 π)

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