What is the integral of 4/(sqrt(x^2+2x+5)) ?

The integral of 4/(sqrt(x^2+2x+5)) is 4ln(1/2 |x+1+sqrt(x^2+2x+5)|)+C

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integral of 4/(sqrt(x^2+2x+5))

\int \frac{4}{x ^{2} + 2 x + 5} d x

4 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5) + C

Solution steps

\int \frac{4}{x ^{2} + 2 x + 5} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} + 2 x + 5} d x

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= 4 \cdot \int sec (v) d v

Use the common integral:

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

= 4 ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

4 ln tan (arctan (\frac{1}{2} (x + 1))) + sec (arctan (\frac{1}{2} (x + 1))) : 4 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

= 4 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

Add a constant to the solution

= 4 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5) + C

y (1 + x^{2}) y^{^{'}} - x (8 + y^{2}) = 0

\frac{d}{d x} (\frac{6}{x} - \frac{8}{x})

a^{x}

\frac{d}{d x} (- cos (4 x))

\frac{\partial}{\partial x} (ln (x^{2} + x))

What is the integral of 4/(sqrt(x^2+2x+5)) ?
The integral of 4/(sqrt(x^2+2x+5)) is 4ln(1/2 |x+1+sqrt(x^2+2x+5)|)+C