# How do you find #int (x+2)/sqrt(-x^2-4x)#?

See the answer below:

By signing up, you agree to our Terms of Service and Privacy Policy

To find the integral of (x + 2)/sqrt(-x^2 - 4x), you can use trigonometric substitution. Let ( t = -x - 2 ). Then ( dt = -dx ). Substituting these into the integral, you get [ \int{\frac{-(t - 2)}{\sqrt{t^2}}} , dt ] Simplify the expression to [ -\int{\frac{t - 2}{\sqrt{t^2}}} , dt ] Now, split the fraction and integrate each part separately. [ -\int{\frac{t}{\sqrt{t^2}}} , dt + 2\int{\frac{1}{\sqrt{t^2}}} , dt ] [ = -\int{\frac{t}{|t|}} , dt + 2\int{\frac{1}{|t|}} , dt ] [ = -\int{sgn(t)} , dt + 2\int{\frac{1}{|t|}} , dt ] The integral of ( sgn(t) ) is ( |t| ) and the integral of ( \frac{1}{|t|} ) is ( ln|t| ). So, the integral becomes [ = -|t| + 2ln|t| + C ] Substituting back ( t = -x - 2 ) gives [ = -|-x - 2| + 2ln|-x - 2| + C ] [ = x + 2 + 2ln|-x - 2| + C ]

By signing up, you agree to our Terms of Service and Privacy Policy

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7