Find the Indefinite Integral using integration by substitution: `int(2x)/(3x+1) dx` this is what I have: u=3x+1 and x=(u-1)/3 then du/dx = 3 and dx= 1/3 du S2(1/3)(u-1)/u (1/3)du 2/9 S (u-1)/u...
Find the Indefinite Integral using integration by substitution: `int(2x)/(3x+1) dx`
this is what I have: u=3x+1 and x=(u-1)/3 then du/dx = 3 and dx= 1/3 du
S2(1/3)(u-1)/u (1/3)du
2/9 S (u-1)/u du
2/9 S (1-1)/u du
2/9 S u-lnu
2/9(3x+1-ln|3x+1|)+C
Thank you for your assistance
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`int (2x)/(3x+1)dx`
Express the denominator with one term only. So, using u-substitution, let the denominator be
`u=3x+1`
Differentiate u with respect to x.
`du=3dx`
`dx=(du)/3`
Since the numerator has a variable x, from `u=3x+1` , solve for x.
`u=3x+1`
`u-1=3x`
`1/3(u-1)=x`
Replacing the x in the integrand with u yields:
`int( 2(1/3u-1))/u *(du)/3=2/9int (u-1)/u du`
Express the integrand as two fractions.
`=2/9 int (u/u-1/u) du = 2/9 int (1-1/u)du= 2/9 int du - 2/9int (du)/u`
`= 2/9u - 2/9 ln u + C`
To return to the original variable, substitute u= 3x +1.
`=2/9 (3x+1) - 2/9 ln(3x+1) + C`
`=2/3x+2/9-2/9 ln(3x+1) + C`
Hence,` int (2x)/(3x+1) dx= 2/3x+2/9-2/9ln(3x+1)+C` .
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