Please help!

∫17dx/x^2(sqrt(x^2+49)) .. let x=7tan(θ).

thank you so much for your help!

To solve this integral, we can start by substituting x = 7tan(θ) into the integral. Let's go step by step.

1. First, let's find dx in terms of dθ using the derivative of the tangent function:

x = 7tan(θ)
dx = 7sec^2(θ)dθ

2. Next, let's substitute x and dx into the integral:

∫ 17dx / x^2(sqrt(x^2+49)) = ∫ 17(7sec^2(θ)dθ) / (7tan^2(θ))(sqrt((7tan(θ))^2+49))

Simplifying, we have:
= (17/7) ∫ (sec^2(θ)dθ) / (tan^2(θ))(sqrt(49tan^2(θ)+49))

3. Now, let's simplify further inside the square root:

sqrt(49tan^2(θ)+49) = sqrt(49(tan^2(θ)+1))
= 7sqrt(tan^2(θ)+1)
= 7sec(θ)

4. Substituting this simplified expression back into the integral, we have:

= (17/7) ∫ (sec^2(θ)dθ) / (tan^2(θ))(7sec(θ))
= (17/7) ∫ (1/cos^2(θ)) / (sin^2(θ))(7)
= (17/7) ∫ (1/(cos^2(θ)sin^2(θ))) dθ

5. Now, let's simplify the denominator:

cos^2(θ)sin^2(θ) = (cos(θ)sin(θ))^2
= (sin(2θ)/2)^2
= sin^2(2θ)/4

6. Substituting this simplified expression back into the integral, we have:

= (17/7) ∫ (1/(sin^2(2θ)/4)) dθ
= (17/7) ∫ (4/sin^2(2θ)) dθ
= (68/7) ∫ (1/sin^2(2θ)) dθ

7. Finally, integrate (1/sin^2(2θ)):

∫ (1/sin^2(2θ)) dθ = -cot(2θ)

8. Substituting the integral back into the original equation, we have:

∫17dx/x^2(sqrt(x^2+49)) = (68/7) ∫ -cot(2θ) dθ

Remember that we initially made the substitution x = 7tan(θ). To complete the solution, you need to integrate -(68/7)cot(2θ) with respect to θ. Once you find the integral, you can convert it back to x by substituting θ with the appropriate expression involving x.