Or use a hyperbolic trig sub where x=(sqrt7)sinh(t).....using 1+sinh^2=cosh^2, the integral just becomes integral(dt), which is of course t. Where t = arcsinh(x/sqrt7) which is a perfectly valid answer. If you want to express the arcsinh as a more elementary function than arcsinh (t) = ln|t + sqrt(1+t^2)|
![LISA - ROCKSTAR (Official Music Video) REACTION [SHE'S BACK!]](http://i.ytimg.com/vi/QBxE30vavMQ/mqdefault.jpg)
Or use a hyperbolic trig sub where x=(sqrt7)sinh(t).....using 1+sinh^2=cosh^2, the integral just becomes integral(dt), which is of course t.
Where t = arcsinh(x/sqrt7) which is a perfectly valid answer.
If you want to express the arcsinh as a more elementary function than arcsinh (t) = ln|t + sqrt(1+t^2)|