one doubt
the integration of log x DX=
Replies (4)
x (log x -1) + c
where c is constant
where c is constant
explain the answer please
Let ∫logxdx=∫(logx*1)dx
(* is multiplication)
Now, by using formula
∫uvdx=u∫vdx-∫[du/dx*∫vdx]dx
Let u=logx and v=1
∫logxdx=logx∫dx-∫[d/dx(logx)*∫dx]dx
=logx*x-∫(1/x*x)dx
=xlogx-∫dx
=xlogx-x+C
=x(log -1) +c
(* is multiplication)
Now, by using formula
∫uvdx=u∫vdx-∫[du/dx*∫vdx]dx
Let u=logx and v=1
∫logxdx=logx∫dx-∫[d/dx(logx)*∫dx]dx
=logx*x-∫(1/x*x)dx
=xlogx-∫dx
=xlogx-x+C
=x(log -1) +c
above answer is crct
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