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

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

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integral of sqrt(x(x+1/x))

\int x (x + \frac{1}{x}) d x

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

Solution steps

\int x (x + \frac{1}{x}) d x

Expand

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

= \int x^{2} + 1 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

f (x) = 41 x - 16 x^{2}, at x = 1

\int \frac{1}{1 + sin ( x ) cos ( x )} d x

(1 + x)^{- \frac{6}{5}}

\int t sin (π t) d t

y = (1 - 3 x)^{10}, (0, 1)

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