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

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

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integral of sqrt(1+(x)^2)

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

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

Solution steps

\int 1 + x^{2} d x

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 (x)

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

Simplify

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

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

Add a constant to the solution

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

\int (x + 2) (x - 6) d x

9 y^{^{''}} - 6 y^{^{'}} + 15 y = 0

\int sin (10 x) cos (7 x) d x

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