What is the integral of y(1+sqrt(3+y^2)) ?

The integral of y(1+sqrt(3+y^2)) is 3/2+1/2 y^2+1/3 (3+y^2)^{3/2}+C

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integral of y(1+sqrt(3+y^2))

\int y (1 + 3 + y^{2}) d y

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

Solution steps

\int y (1 + 3 + y^{2}) d y

Apply u-substitution

= \int \frac{1 + u}{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 1 + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d u + \int u d u)

\int 1 d u = u

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{2} (u + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 3 + y^{2}

= \frac{1}{2} (3 + y^{2} + \frac{2}{3} (3 + y^{2})^{\frac{3}{2}})

Simplify

\frac{1}{2} (3 + y^{2} + \frac{2}{3} (3 + y^{2})^{\frac{3}{2}}) : \frac{3}{2} + \frac{1}{2} y^{2} + \frac{1}{3} (3 + y^{2})^{\frac{3}{2}}

= \frac{3}{2} + \frac{1}{2} y^{2} + \frac{1}{3} (3 + y^{2})^{\frac{3}{2}}

Add a constant to the solution

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

What is the integral of y(1+sqrt(3+y^2)) ?
The integral of y(1+sqrt(3+y^2)) is 3/2+1/2 y^2+1/3 (3+y^2)^{3/2}+C