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

The integral of sqrt(1+u^2) is 1/2 usqrt(1+u^2)+1/2 ln|u+sqrt(1+u^2)|+C

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integral of sqrt(1+u^2)

\int 1 + u^{2} d u

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

Solution steps

\int 1 + u^{2} d u

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

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

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

Substitute back

v = arctan (u)

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{4} x^{3} x^{2} + 16 d x

\frac{d}{d x} (3 x^{2} + 2 x + 1)

t^{2} y^{^{'}} + t y = 5

x sin (3 x)

\frac{d}{d x} (\frac{1}{e ^{x} + 1})

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