Best Answer

This is solved by using substitution:

Let x = sin θ

→ dx = cos θ dθ

and √(1 - x²) = √(1 - sin² θ) = cos θ

and θ = arcsin x

→ ∫ (x + 1)/√(1 - x²) dx

= ∫ ((sin θ + 1)/ cos θ) cos θ dθ

= ∫ sin θ + 1 dθ

= -cos θ + θ + b

= θ - cos θ + b

= arcsin x - √(1 - x²) + c

Q: How do you answer the equation integral of x 1 over square root of 1-x2 dx?

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