To evaluate the given integral problem:` int 2/(7e^x+4)dx` , we may apply u-substitution using: `u= e^x ` then `du = e^x dx` .
Plug-in `u = e^x` on `du= e^x dx` , we get: `du = u dx` or `(du)/u =dx`
The integral becomes:
`int 2/(7e^x+4)dx =int 2/(7u+4)* (du)/u`
`=int 2/(7u^2+4u)du`
Apply the basic properties of integration:` int c*f(x) dx= c int f(x)...
To evaluate the given integral problem:` int 2/(7e^x+4)dx` , we may apply u-substitution using: `u= e^x ` then `du = e^x dx` .
Plug-in `u = e^x` on `du= e^x dx` , we get: `du = u dx` or `(du)/u =dx`
The integral becomes:
`int 2/(7e^x+4)dx =int 2/(7u+4)* (du)/u`
`=int 2/(7u^2+4u)du`
Apply the basic properties of integration:` int c*f(x) dx= c int f(x) dx` .
`int 2/(7u^2+4u)du =2int 1/(7u^2+4u)du`
Apply completing the square: `7u^2+4u =(sqrt(7)u+2/sqrt(7))^2 -4/7`
`2int 1/(7u^2+4u)du =2int 1/((sqrt(7)u+2/sqrt(7))^2 -4/7)du`
Let `v =sqrt(7)u+2/sqrt(7)` then `dv = sqrt(7) du` or `(dv)/sqrt(7) = du` .
The integral becomes:
`2int 1/(7u^2+4u)du =2 int 1/(v^2 -4/7) *(dv)/sqrt(7)`
Rationalize the denominator:
`2 int 1/(v^2 -4/7) *(dv)/sqrt(7) *sqrt(7)/sqrt(7)`
`= 2 int (sqrt(7)dv)/ ( 7*(v^2 -4/7))`
`=2 int (sqrt(7)dv)/ ( 7v^2 -4)`
From the table of integrals, we may apply `int dx/(x^2-a^2) = 1/(2a)ln[(u-a)/(u+a)]+C`
Let:` w = sqrt(7)v` then `dw = sqrt(7) dv`
`2int (sqrt(7) dv)/ ( 7v^2 -4) =2int (sqrt(7) dv)/ (( sqrt(7)v)^2 -2^2)`
`= 2 int (dw)/ (w^2-2^2)`
`= 2 *1/(2*2)ln[(w-2)/(w+2)]+C`
`=1/2ln[(w-2)/(w+2)]+C`
Recall we let: `w =sqrt(7)v` and `v =sqrt(7)u+2/sqrt(7)` .
Then, `w=sqrt(7)*[sqrt(7)u+2/sqrt(7)] = 7u +2`
Plug-in `u =e^x` on `w=7u +2` , we get: `w= 7e^x+2`
The indefinite integral will be:
`int 2/(7e^x+4)dx =1/2ln[(7e^x+2-2)/(7e^x+2+2)]+C`
` =1/2ln[(7e^x)/(7e^x+4)]+C`
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