What is the integral of (sqrt(x^2+4x+3))/(x+2) ?

The integral of (sqrt(x^2+4x+3))/(x+2) is -arctan(sqrt(x^2+4x+3))+sqrt(x^2+4x+3)+C

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integral of (sqrt(x^2+4x+3))/(x+2)

\int \frac{x ^{2} + 4 x + 3}{x + 2} d x

- arctan (x^{2} + 4 x + 3) + x^{2} + 4 x + 3 + C

Solution steps

\int \frac{x ^{2} + 4 x + 3}{x + 2} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{v ^{2}}{v ^{2} + 1} d v

= \int \frac{v ^{2}}{v ^{2} + 1} d v

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

= \int - \frac{1}{v ^{2} + 1} + 1 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

= - \int \frac{1}{v ^{2} + 1} d v + \int 1 d v

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= - arctan (v) + v

Substitute back

= - arctan ((x + 2)^{2} - 1) + (x + 2)^{2} - 1

Simplify

- arctan ((x + 2)^{2} - 1) + (x + 2)^{2} - 1 : - arctan (x^{2} + 4 x + 3) + x^{2} + 4 x + 3

= - arctan (x^{2} + 4 x + 3) + x^{2} + 4 x + 3

Add a constant to the solution

= - arctan (x^{2} + 4 x + 3) + x^{2} + 4 x + 3 + C

\int e^{x} 49 + e^{x} d x

y^{^{''}} + 2 y = - 18 x^{2} e^{2 x}

\int_{1}^{0} - 9 x d x

y^{^{''}} - 4 y^{^{'}} - 12 y = t e^{4 t}

e^{x} \frac{( 2 x + 5 ) ^{2}}{x - 4}

What is the integral of (sqrt(x^2+4x+3))/(x+2) ?
The integral of (sqrt(x^2+4x+3))/(x+2) is -arctan(sqrt(x^2+4x+3))+sqrt(x^2+4x+3)+C