What is the integral of (x^2+1)^{0.5} ?

The integral of (x^2+1)^{0.5} is 1/2 xsqrt(x^2+1)+1/2 ln|x+sqrt(1+x^2)|+C

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integral of (x^2+1)^{0.5}

integral

(x^{2} + 1)^{0.5}

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

Solution steps

\int (x^{2} + 1)^{0.5} d x

Simplify

x^{2} : xx

= \int (xx + 1)^{0.5} d x

xx = x^{2}

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

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

Add a constant to the solution

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

\frac{1}{( x + a )}

(ln (x))^{3}

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

\frac{ln ( t )}{( t + 1 ) ^{2}}

\frac{x ^{2}}{( x ^{2} - 2 x + 2 ) ^{2}}

What is the integral of (x^2+1)^{0.5} ?
The integral of (x^2+1)^{0.5} is 1/2 xsqrt(x^2+1)+1/2 ln|x+sqrt(1+x^2)|+C