Q. `int` `log(x+1) dx` =?

A) `xlogx - x + log(x+1) + c`

B) `(x+1)log(x+1) - x +c`

C) `log(x/(x+1)) + c`

D) `log [(1+x)/(2+x)] + c`

### 1 Answer | Add Yours

`intlog(x+1)dx`

This can be done using integral by parts.

Let;

`u = log(x+1)`

`du = 1/(x+1)dx`

`v = x`

`dv = dx`

`intudv = uv-intvdu`

`intlog(x+1)dx = xlog(x+1)-intx/(x+1)dx`

`intlog(x+1)dx = xlog(x+1)-int(x+1-1)/(x+1)dx`

`intlog(x+1)dx = xlog(x+1)-[int(x+1)/(x+1)dx-int1/(x+1)dx]`

`intlog(x+1)dx = xlog(x+1)-intdx+int1/(x+1)dx`

`intlog(x+1)dx = xlog(x+1)-x+log(x+1)+C` where C is a constant.

`intlog(x+1)dx = (x+1)log(x+1)-x+C`

** So the answer is;**`intlog(x+1)dx = (x+1)log(x+1)-x+C`

*Correct answer is B.*

**Sources:**

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