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

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

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integral of sqrt(1+y^2)

\int 1 + y^{2} d y

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

Solution steps

\int 1 + y^{2} 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

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

\frac{( x ^{4} + 1 )}{e ^{x} + 2 x + 1}

12 x^{3} - 24 x^{2}

f (I) = I^{22}

\int \frac{x ^{2} + 8}{x ^{4} - x ^{3} - 2 x ^{2}} d x

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