intergral of:

log(3x+2)dx

log(3x+2)dxLet 3x+2=t3dx=dtdx=dt/3putting valuesdt3log t=13logt dt=13[logtdt-[dlogtdtdt] dt=13[tlogt-1t×t dt]=13[tlogt-t]=t3[logt-1]Putting value of t=(3x+2)3[log(3x+2)-1]=(3x+2)3[log(3x+2)-log e]=(3x+2)3[log (3x+2)e] answer

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Let 3x+2=t

So, dx=dt/3

So, integral log(3x+2)dx = 1/3(integral logtdt)

or integral log(3x+2)dx = 1/3(integral logt.1dt)

Integrating by parts we get,

integral log(3x+2)dx = 1/3[tlogt - integral t(1/t)dt]

or integral log(3x+2)dx = 1/3[tlogt - integral dt]

or integral log(3x+2)dx = 1/3(tlogt - t)

or integral log(3x+2)dx = 1/3[t(logt - 1)]

or integral log(3x+2)dx = 1/3[t(logt - loge)]

or integral log(3x+2)dx = 1/3tlog(t/e)

or integral log(3x+2)dx = 1/3(3x+2)log[(3x+2)/e].

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