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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

= \int u^{2} + 1 d u

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

= \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) ∣

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((9 x + 2 y)^{5})

\int_{t - 1}^{t} (1 - t + x) \cdot x d x

\int_{0}^{\frac{1}{4}} 2 arcsin (4 y) d y

\int sin (\frac{π x ^{2}}{2}) d x

f (x) = \frac{x + 3}{x - 3}

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