cyphactor | And if I am understanding this correctly it is a vector which is perpendicular to the two planes because I am taking the normal vectors of the two planes and cross-producting them which gives me another vector which is perpendicular (normal) to the original two planes normal vectors. Hence, when I use that resulting normal from the cross-product I get a plane which is perpendicular to that normal. Hence, perpendicular two the two original planes Does that make sense? |

Phylo | where do I begin to solve Integral[1/(x^2 * sqr(1+x^2))] split fractions? *partial fractions8 or trigonometric substitution? |

excelblue | try trig substitution try what you think may work, and do it if you seem stuck, try another method or even try the technique of letting u = the square root part (eg 1/sqrt(1+x^2)) |

Phylo | integration by parts? nah, that was last chapter, it probably won't work I'm thinking trig. sub. but I don't see what to do... |

lieven | try x=tan u |

Phylo | will try: what would make the denominator a quadratic denominator oh, non-linear? % Integrate[Cotan(x), x] |

mbot | Phylo: (Cotan*x^2)/2 |

Phylo | % Integrate[Cotan(x)^2, x] |

mbot | Phylo: (Cotan*x^3)/3 |

|Steve| | % Integrate[Cot[x],x] |

mbot | |Steve|: Log[Sin[x]] |

Phylo | ... |

|Steve| | Mathematica can't read your mind. |

Phylo | thx, I realized it when I saw the Cotan"*"x^2 % Integrate[Cot[x]^2, x] |

mbot | Phylo: -x - Cot[x] |

thermoplyae | cyphactor: Sorry I went idle for so long. That's all correct. And on that note, goodnight |

Phylo | I substitued tan(u) for x now I have Integrate[1/(Tan(u)^2), u] but I have to convert back to x hmm % Integrate[1/(Tan(u)^2), u] |

mbot | Phylo: -(1/(Tan*u)) |

Phylo | % Integrate[1/(Tan[u]^2), u] |

mbot | Phylo: -u - Cot[u] |

lieven | phylo: I think you've made a mistake in working it out |

Phylo | somewhere % Integrate[1/(x^2*SQR[1+x^2]), x] |

mbot | Phylo: Integrate[1/(x^2*SQR[1 + x^2]), x] |