# For what value of C  `int_0^c ln (x +1)dx=1`

sciencesolve | Certified Educator

You need to use integration by parts to evaluate the given integral, such that:

`ln(x+1) = u => 1/(x+1)dx = du`

`dx = dv => x = v`

`int_0^c ln(x+1)dx = x*ln(x+1)|_0^c - int_0^c x/(x + 1)dx`

You need to evaluate the definite integral `int_0^c x/(x+1)dx` such that:

`int_0^c x/(x + 1)dx = int_0^c (x + 1 - 1)/(x + 1)dx`

Using the property of linearity of integral yields:

`int_0^c x/(x + 1) = int_0^c (x + 1)/(x + 1)dx -int_0^c 1/(x + 1)dx`

`int_0^c x/(x + 1) = x|_0^c - ln(x+1)|_0^c`

`int_0^c ln(x+1)dx = x*ln(x+1)|_0^c - x|_0^c + ln(x+1)|_0^c`

Using the fundamental theorem of calculus yields:

`int_0^c ln(x+1)dx = c*ln(c+1) - 0*ln 1 - c + 0 + ln(c+1) - ln 1`

Since `ln 1 = 0` yields:

`int_0^c ln(x+1)dx = c*ln(c+1) - c + ln(c+1)`

You need to evaluate c, hence, you need to use the information provided by the problem, such that:

`int_0^c ln(x+1)dx = 1 => c*ln(c+1) - c + ln(c+1) = 1`

`c*ln(c+1) - c = 1 - ln(c+1)`

You need to factor out c to the left side, such that:

`c(ln(c+1) - 1) = -(ln(c+1) - 1)`

Reducing duplicate factors both sides, yields:

`c = -1`

Hence, evaluating the value of c, under the given condition, yields `c = -1.`