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

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

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integral of sqrt(y^2+1)

\int y^{2} + 1 d y

\frac{1}{2} y 1 + y^{2} + \frac{1}{2} ln y + 1 + y^{2} + C

Solution steps

\int y^{2} + 1 d y

Apply Trigonometric Substitution

: \int sec^{3} (u) d u

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

Apply Integral Reduction

= \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) ∣

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

Substitute back

u = arctan (y)

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

Simplify

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

= \frac{1}{2} y 1 + y^{2} + \frac{1}{2} ln y + 1 + y^{2}

Add a constant to the solution

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

y = sin (sin (x)), at x = π

y = x, (9, 3)

\frac{sin ( x ) + cos ( x )}{cos ( x )}

y = \frac{8 x}{x ^{2} + 1}, (- 1, - 4)

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