Evaluate integral (3x-1)/(x+2)2 dx

3x-1(x+2)2dxLet x+2 =tso dx =dtand 3(t-2)-1 =3t-7So the integration get changed into3t-7t2dt=3tt2-7t2dt=3t-7t2dt= 3ln(t) +7t+C=3ln(x+2) +7x+2+C

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Let x + 2 = t

So, x = t - 2

or 3x - 1 = 3t - 7

Also,dx = dt.

So integral (3x-1)/(x+2)2dx = integral (3t-7)/t2dt

or integral (3x-1)/(x+2)2dx = integral 3t/t2dt - integral 7/t2dt

or integral (3x-1)/(x+2)2dx = integral 3/tdt - 7integral 1/t2dt

or integral (3x-1)/(x+2)2dx = 3integral 1/tdt - 7integral t-2dt

or integral (3x-1)/(x+2)2dx = 3log|t| - 7(-t-1)

or integral (3x-1)/(x+2)2dx = 3log|t| + 7/t

or integral (3x-1)/(x+2)2dx = 3log|x+2| + 7/(x+2).

Hence, integral (3x-1)/(x+2)2dx = 3log|x+2| + 7/(x+2).

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