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intlog(x+sqrt(x^2+a^2))dx...

`intlog(x+sqrt(x^2+a^2))dx`

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`"Let I"=int log(x+sqrt(x^(2)+a^(2)))dx`
`=int log (x+sqrt(x^(2)+a^(2))).1dx`
`=[log (x+sqrt(x^(2)+a^(2))].int 1dx-int {(d)/(dx)[log x+sqrt((x^(2)+a^(2)))[int 1 dx}dx`
`"Now", (d)/(dx)[log (x+sqrt(x^(2)+a^(2))]=(1)/(x+sqrt(x^(2)+a^(2))).(d)/(dx)[x+sqrt(x^(2)+a^(2))]`
`=(1)/(x+sqrt(x^(2)+a^(2))).[1+(1)/(2sqrt(x^(2)+a^(2))).(d)/(dx)(x^(2)+a^2)]`
`=(1)/(x+sqrt(x^(2)+a^(2))).[1+(1)/(2sqrt(x^(2)+a^(2))).2x]`
`=(1)/(x+sqrt(x^(2)+a^(2))).[(sqrt(x^(2)+a^(2)+x))/(sqrt(x^(2)+a^(2)))]`
`=(1)/(sqrt(x^(2)+a^(2)))`
`therefore I=[log(x+sqrt(x^(2)+a^(2))].(x)-int (1)/(sqrt(x^(2)+a^(2))).xdx`
`=x.log(x+sqrt(x^(2)+a^(2)))-(1)/(2)int (2x)/(sqrt(x^(2)+a^(2)))dx`
`"Put "x^(2)+a^(2)=t" "therefore 2xdx=dt`
`therefore I=x.log(x+sqrt(x^(2)+a^(2)))-(1)/(2)int (1)/(sqrtt)dt`
`=x.log (x+sqrt(x^(2)+a^(2)))-(1)/(2)int t^(-(1)/(2))dt`
`=x.log(x+sqrt(x^(2)+a^(2)))-(1)/(2).(t^((1)/(2)))/(((1)/(2)))+c`
`=x.log(x+sqrt(x^(2)+a^(2)))-sqrt(x^(2)+a^(2))+c`
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