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

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

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

\int 1 + y^{2} 2 y d y

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

Solution steps

\int 1 + y^{2} \cdot 2 y d y

Take the constant out:

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

= 2 \cdot \int 1 + y^{2} y d y

Apply u-substitution

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

Apply the Power Rule

= 2 \cdot \frac{u ^{3}}{3}

Substitute back

u = 1 + y^{2}

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

Simplify

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

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

Add a constant to the solution

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

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