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

The integral of sqrt(x^2+1x) is 1/8 (4xsqrt(x(x+1))+2sqrt(x(x+1))-ln|2x+1+2sqrt(x(x+1))|)+C

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integral of sqrt(x^2+1x)

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

\frac{1}{8} (4 x x (x + 1) + 2 x (x + 1) - ln 2 x + 1 + 2 x (x + 1)) + C

Solution steps

\int x^{2} + 1 \cdot x d x

Simplify

= \int x^{2} + x d x

Complete the square

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

= \int (x + \frac{1}{2})^{2} - \frac{1}{4} d x

Apply u-substitution

: \int \frac{4 u ^{2} - 1}{2} d u

= \int \frac{4 u ^{2} - 1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 4 u^{2} - 1 d u

Apply Trigonometric Substitution

: \int \frac{1}{2} tan^{2} (v) sec (v) d v

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

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

Expand

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int - sec (v) + sec^{3} (v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot \frac{1}{2} (- \int sec (v) d v + \int sec^{3} (v) d v)

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

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} \cdot \frac{1}{2} (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} (- ln tan (arcsec (2 (x + \frac{1}{2}))) + sec (arcsec (2 (x + \frac{1}{2}))) + \frac{1}{2} sec (arcsec (2 (x + \frac{1}{2}))) tan (arcsec (2 (x + \frac{1}{2}))) + \frac{1}{2} ln tan (arcsec (2 (x + \frac{1}{2}))) + sec (arcsec (2 (x + \frac{1}{2}))))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- ln tan (arcsec (2 (x + \frac{1}{2}))) + sec (arcsec (2 (x + \frac{1}{2}))) + \frac{1}{2} sec (arcsec (2 (x + \frac{1}{2}))) tan (arcsec (2 (x + \frac{1}{2}))) + \frac{1}{2} ln tan (arcsec (2 (x + \frac{1}{2}))) + sec (arcsec (2 (x + \frac{1}{2})))) : \frac{1}{8} (4 x x (x + 1) + 2 x (x + 1) - ln 2 x + 1 + 2 x (x + 1))

= \frac{1}{8} (4 x x (x + 1) + 2 x (x + 1) - ln 2 x + 1 + 2 x (x + 1))

Add a constant to the solution

= \frac{1}{8} (4 x x (x + 1) + 2 x (x + 1) - ln 2 x + 1 + 2 x (x + 1)) + C

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What is the integral of sqrt(x^2+1x) ?
The integral of sqrt(x^2+1x) is 1/8 (4xsqrt(x(x+1))+2sqrt(x(x+1))-ln|2x+1+2sqrt(x(x+1))|)+C