Question

Is my solution correct?
Pic attached

Answer 1

The correct one is the third

Answer 2

∫e^{t}√(169-e^{2t})dt
e^{t}=x⇒e^{t}dt=dx
∫√(169-x²)dx
I=∫x′√(169-x²)dx=x√(169-x²)+2∫((x²-169+169)/(2√(169-x²)))dx=
=x√(169-x²)-I+∫((169)/(√(169-x²)))dx
I=(1/2)x√(169-x²)+((169)/2)arcsin(x/(13))
∫((169)/(√(169-x²)))dx=169arcsin(x/(13))