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

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

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integral of ysqrt(1+y^2)

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

x^{^{''}} - 5 x^{^{'}} + 6 x = (t^{2} + t) sin (t)

x \to - 1 - lim (g (x))

f (x) = \frac{3}{( 2 x + 1 ) ^{\frac{5}{2}}}

x y^{^{'}} - y = - x^{2} + 2 x + 2

\int \frac{x ^{2} + 1}{x ^{2} - x} d x

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