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

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

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integral of ysqrt(y^2+1)

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

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

Substitute back

u = y^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

x \to 3 lim (2 x + 4)

n = 0 \sum \infty 2 \cdot 0. 3^{n}

\frac{d y}{d x} + x (y^{2} + y) = 0

\int \frac{2 tan ( x )}{1 - tan ^{2} ( x )} d x

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