What is the integral of 2/((x+2)^2sqrt(x^2+4x+13)) ?

The integral of 2/((x+2)^2sqrt(x^2+4x+13)) is -(2sqrt(x^2+4x+13))/(9(x+2))+C

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integral of 2/((x+2)^2sqrt(x^2+4x+13))

\int \frac{2}{( x + 2 ) ^{2} x ^{2} + 4 x + 13} d x

- \frac{2 x ^{2} + 4 x + 13}{9 ( x + 2 )} + C

Solution steps

\int \frac{2}{( x + 2 ) ^{2} x ^{2} + 4 x + 13} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{( x + 2 ) ^{2} x ^{2} + 4 x + 13} d x

Apply u-substitution

: \int \frac{1}{u ^{2} u ^{2} + 9} d u

= 2 \cdot \int \frac{1}{u ^{2} u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{9 tan ^{2} ( v )} d v

= 2 \cdot \int \frac{sec ( v )}{9 tan ^{2} ( v )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \int \frac{sec ( v )}{tan ^{2} ( v )} d v

Rewrite using trig identities

= 2 \cdot \frac{1}{9} \cdot \int \frac{cos ( v )}{sin ^{2} ( v )} d v

Apply u-substitution

: \int \frac{1}{w ^{2}} d w

= 2 \cdot \frac{1}{9} \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

: - \frac{1}{w}

= 2 \cdot \frac{1}{9} (- \frac{1}{w})

Substitute back

= 2 \cdot \frac{1}{9} (- \frac{1}{sin ( arctan ( \frac{1}{3} ( x + 2 ) ) )})

Simplify

2 \cdot \frac{1}{9} (- \frac{1}{sin ( arctan ( \frac{1}{3} ( x + 2 ) ) )}) : - \frac{2 x ^{2} + 4 x + 13}{9 ( x + 2 )}

= - \frac{2 x ^{2} + 4 x + 13}{9 ( x + 2 )}

Add a constant to the solution

= - \frac{2 x ^{2} + 4 x + 13}{9 ( x + 2 )} + C

\int e^{2 y^{2}} d y

x \to 1 lim (f (x))

\int 60 sec^{4} (x) d x

6 t^{2} + 3 t + 7

y^{^{'}} + 4 y = 12

What is the integral of 2/((x+2)^2sqrt(x^2+4x+13)) ?
The integral of 2/((x+2)^2sqrt(x^2+4x+13)) is -(2sqrt(x^2+4x+13))/(9(x+2))+C