How Can I Evaluate These Integrals Using U-Substitution?

[Ed Quanta]

Sorry I don't know how to use the notation, but help me solve the following through the u substitution.

1) dx/square root of x^2+5x

2)dx/x^2*square root of 1-x^2

I am embarassed that I am not able to solve these still after devoting much thought, but help from anyone will be appreciated.

 

[Tide]

(1) Complete the square

(2) Let u = 1/x

 

[M98Ranger]

No need to be embarrassed, integration can be tricky and it takes practice to become comfortable with it. Let's go through each integral step by step:

1) dx/square root of x^2+5x

To solve this integral, we will use the u-substitution method. Let u = x^2 + 5x, then du = (2x + 5)dx. We can rearrange this to get dx = du/(2x + 5).

Substituting this into the original integral, we get:

∫dx/square root of x^2+5x = ∫du/(2x + 5)

Next, we need to get rid of the x in the denominator. We can do this by factoring out an x from the denominator, giving us:

∫du/(2x + 5) = ∫du/x(2 + 5/x)

Now, we can use the u-substitution again, this time with v = 2 + 5/x. This gives us dv = -5/x^2 dx. Rearranging this, we get dx = -5/(v^2 - 5)dv.

Substituting this into the integral, we get:

∫du/x(2 + 5/x) = ∫-5/(v^2 - 5)dv

Now, we can use a trigonometric substitution to solve this integral. Let v = √5secθ, then dv = √5tanθsecθ dθ. Substituting this into the integral, we get:

∫-5/(v^2 - 5)dv = ∫-5/(5sec^2θ - 5)√5tanθsecθ dθ

Simplifying this, we get:

∫-5/(v^2 - 5)dv = ∫-√5tanθ dθ

Using the trigonometric identity tan^2θ + 1 = sec^2θ, we can rearrange this to get:

∫-√5tanθ dθ = ∫-√5(tan^2θ + 1 - 1) dθ

= ∫-√5(sec^2θ - 1) dθ = ∫-√5(v^2/5 - 1) dθ

= ∫-