What is the integral of sqrt(x^2+6x+13) ?

The integral of sqrt(x^2+6x+13) is 4(1/8 (x+3)sqrt(x^2+6x+13)+1/2 ln((|x+3+sqrt(x^2+6x+13)|)/2))+C

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integral of sqrt(x^2+6x+13)

\int x^{2} + 6 x + 13 d x

4 (\frac{1}{8} (x + 3) x^{2} + 6 x + 13 + \frac{1}{2} ln (\frac{x + 3 + x ^{2} + 6 x + 13}{2})) + C

Solution steps

\int x^{2} + 6 x + 13 d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

= \int (x + 3)^{2} + 4 d x

Apply u-substitution

: \int u^{2} + 4 d u

= \int u^{2} + 4 d u

Apply Trigonometric Substitution

: \int 4 sec^{3} (v) d v

= \int 4 sec^{3} (v) d v

Take the constant out:

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

= 4 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 4 (\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) ∣

= 4 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} ( x + 3 ) ) ) sin ( arctan ( \frac{1}{2} ( x + 3 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} (x + 3))) + sec (arctan (\frac{1}{2} (x + 3))))

Simplify

4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} ( x + 3 ) ) ) sin ( arctan ( \frac{1}{2} ( x + 3 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} (x + 3))) + sec (arctan (\frac{1}{2} (x + 3)))) : 4 (\frac{1}{8} (x + 3) x^{2} + 6 x + 13 + \frac{1}{2} ln (\frac{x + 3 + x ^{2} + 6 x + 13}{2}))

= 4 (\frac{1}{8} (x + 3) x^{2} + 6 x + 13 + \frac{1}{2} ln (\frac{x + 3 + x ^{2} + 6 x + 13}{2}))

Add a constant to the solution

= 4 (\frac{1}{8} (x + 3) x^{2} + 6 x + 13 + \frac{1}{2} ln (\frac{x + 3 + x ^{2} + 6 x + 13}{2})) + C

f (x) = \frac{1}{x ^{3}}, at x = 1

\frac{\partial}{\partial x} (y^{6} sin (5 x))

arctan (\frac{4}{x})

\frac{d y}{d x} + y = \frac{e ^{x}}{y}

\int_{3}^{4} e^{3 - x} d x

What is the integral of sqrt(x^2+6x+13) ?
The integral of sqrt(x^2+6x+13) is 4(1/8 (x+3)sqrt(x^2+6x+13)+1/2 ln((|x+3+sqrt(x^2+6x+13)|)/2))+C