solve a integral without using without using trigo-substitution?

**afeasfaerw23231233** · November 2nd 2007, 12:47 PM

**TKHunny** · November 2nd 2007, 01:05 PM

Sure! Look it up in a table.

Really, my first question would be, "Why?"

**Plato** · November 2nd 2007, 01:05 PM

Both of those are in arctangent form.
Look at the derivative of the arctangent(x).

**Krizalid** · November 2nd 2007, 01:49 PM

I think it's an useless question, 'cause it's an immediate integral.

But you can say $y=\arctan x\implies\tan y=x\,\therefore\,y'(1+x^2)=1.$

So, $y'=\frac1{1+x^2}.$

Now integrate both sides $y+k=\int\frac1{1+x^2}\,dx.$

Since $y=\arctan x,$ we happily get that $\arctan x+k=\int\frac1{1+x^2}\,dx.$

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