Question

The indefinite integral $\int \frac{dx}{(x^2+4x+5)^2}$ equals:

Answer 1

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

Answer 2

To evaluate the indefinite integral $\int \frac{dx}{(x^2+4x+5)^2}$, we first complete the square in the denominator:

$x^2 + 4x + 5 = (x^2 + 4x + 4) + 1 = (x+2)^2 + 1$.

The integral becomes $\int \frac{dx}{((x+2)^2+1)^2}$.

Let $u = x+2$. Then $du = dx$. The integral transforms to $\int \frac{du}{(u^2+1)^2}$.

This is a standard integral. We can solve this using trigonometric substitution. Let $u = \tan \theta$. Then $du = \sec^2 \theta \, d\theta$. Also, $u^2+1 = \tan^2 \theta + 1 = \sec^2 \theta$.

The integral becomes $\int \frac{\sec^2 \theta \, d\theta}{(\sec^2 \theta)^2} = \int \frac{\sec^2 \theta}{\sec^4 \theta} \, d\theta = \int \frac{1}{\sec^2 \theta} \, d\theta = \int \cos^2 \theta \, d\theta$.

Using the identity $\cos^2 \theta = \frac{1+\cos(2\theta)}{2}$, the integral is: $\int \frac{1+\cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int (1+\cos(2\theta)) \, d\theta = \frac{1}{2} \left( \theta + \frac{1}{2} \sin(2\theta) \right) + C = \frac{1}{2} \theta + \frac{1}{4} \sin(2\theta) + C$.

Now, we express this in terms of $u$. Since $u = \tan \theta$, we have $\theta = \arctan(u)$. For $\sin(2\theta)$, we use $\sin(2\theta) = 2 \sin \theta \cos \theta$. From $u = \tan \theta = \frac{u}{1}$, we can form a right triangle with opposite side $u$ and adjacent side $1$. The hypotenuse is $\sqrt{u^2+1}$. So, $\sin \theta = \frac{u}{\sqrt{u^2+1}}$ and $\cos \theta = \frac{1}{\sqrt{u^2+1}}$. $\sin(2\theta) = 2 \left( \frac{u}{\sqrt{u^2+1}} \right) \left( \frac{1}{\sqrt{u^2+1}} \right) = \frac{2u}{u^2+1}$.

Substituting back: $\frac{1}{2} \arctan(u) + \frac{1}{4} \left( \frac{2u}{u^2+1} \right) + C = \frac{1}{2} \arctan(u) + \frac{u}{2(u^2+1)} + C$.

Finally, substitute back $u = x+2$: $\frac{1}{2} \arctan(x+2) + \frac{x+2}{2((x+2)^2+1)} + C = \frac{1}{2} \arctan(x+2) + \frac{x+2}{2(x^2+4x+5)} + C$.

Thus, the indefinite integral is $\frac{x+2}{2(x^2+4x+5)} + \frac{1}{2} \arctan(x+2) + C$.