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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

= y \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= y (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= y (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (x)

= y (\frac{sec ^{2} ( arctan ( x ) ) sin ( arctan ( x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (x)) + sec (arctan (x)) ∣)

Simplify

y (\frac{sec ^{2} ( arctan ( x ) ) sin ( arctan ( x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (x)) + sec (arctan (x)) ∣) : y (\frac{1}{2} x (x^{2} + 1)^{\frac{1}{2}} + \frac{1}{2} ln x + 1 + x^{2})

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

Add a constant to the solution

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

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