Math, asked by tulsipareek332, 2 days ago

the spelling is not wrong

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Answered by siddhesh0024

I = integ.of log(2–3x).dx

I= integ.of {log (2–3x) . (1) }.dx

I = {log(2–3x)} × {x} - integ.of { -3/(2–3x)}.x.dx

I= x.log (2–3x) - integ.of [{(2–3x) - 2}/(2–3x)].dx

I = x.log (2–3x) - integ.of [ 1 + (2/3).{ -3/(2–3x)}].dx

I = x.log (2–3x) - [ x + 2/3.log (2–3x)] + C.

I= (x- 2/3).log (2–3x) - x +C.

I = - {(2–3x)/3}.log (2–3x) -x +C. Answer.

I = integ.of log(2–3x).dx

I= integ.of {log (2–3x) . (1) }.dx

I = {log(2–3x)} × {x} - integ.of { -3/(2–3x)}.x.dx

I= x.log (2–3x) - integ.of [{(2–3x) - 2}/(2–3x)].dx

I = x.log (2–3x) - integ.of [ 1 + (2/3).{ -3/(2–3x)}].dx

I = x.log (2–3x) - [ x + 2/3.log (2–3x)] + C.

I= (x- 2/3).log (2–3x) - x +C.

I = - {(2–3x)/3}.log (2–3x) -x +C. Answer.

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I = integ.of log(2–3x).dx

I= integ.of {log (2–3x) . (1) }.dx

I = {log(2–3x)} × {x} - integ.of { -3/(2–3x)}.x.dx

I= x.log (2–3x) - integ.of [{(2–3x) - 2}/(2–3x)].dx

I = x.log (2–3x) - integ.of [ 1 + (2/3).{ -3/(2–3x)}].dx

I = x.log (2–3x) - [ x + 2/3.log (2–3x)] + C.

I= (x- 2/3).log (2–3x) - x +C.

I = - {(2–3x)/3}.log (2–3x) -x +C. Answer.

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