How do you integrate #int 1/sqrt(e^(2x)+12e^x+32)dx# using trigonometric substitution?

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Answer 1 · 2018-03-26T05:36:37

# -sqrt2/8ln|{(16+3e^x)+sqrt(256+96e^x+8e^(2x))}/e^x|+C, or,#

# sqrt2/8[x-ln|{(3e^x+16)+2sqrt2sqrt(e^(2x)+12e^x+32)}|]++C#.

We first use the substitution

#e^x=t," so that, "e^xdx=dt, or, dx=dt/e^x=dt/t#.

#:. I=int1/sqrt(e^(2x)+12e^x+32)dx=int1/{tsqrt(t^2+12t+32)}dt#.

Next, we let, #t=1/y.:. dt=-1/y^2dy#.

Now, #tsqrt(t^2+12t+32)=1/y*sqrt{1/y^2+12/y+32}#,

#=sqrt(1+12y+32y^2)/y^2#.

#:. I=inty^2/sqrt(1+12y+32y^2)*-1/y^2dy#,

#=-int1/sqrt(1+12y+32y^2)dy#,

#=-intsqrt8/sqrt{8(1+12y+32y^2)}dy#,

#=-2sqrt2int1/sqrt(256y^2+96y+8)dy#,

#=-2sqrt2int1/sqrt{(16y+3)^2-1}dy#,

#=-2sqrt2*1/16ln|{(16y+3)+sqrt(256y^2+96y+8)}|#,

#=-sqrt2/8ln|{(16/t+3)+sqrt(256/t^2+96/t+8)}|...[because, y=1/t]#,

#=-sqrt2/8ln|{(16+3t)+sqrt(256+96t+8t^2)}/t|#.

#rArr I=-sqrt2/8ln|{(16+3e^x)+sqrt(256+96e^x+8e^(2x))}/e^x|+C, or,#

#I=sqrt2/8[x-ln|{(3e^x+16)+2sqrt2sqrt(e^(2x)+12e^x+32)}|]++C#.

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