# Evaluate the indefinite integral. integrate of ln(x^2+10x+24)dx

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You need to use integration by parts such that:

`int udv = uv - int vdu`

Considering `u = ln(x^2+10x+24)` and `dv = dx` yields:

`u = ln(x^2+10x+24) => du = (2x+10)/(x^2+10x+24) dx`

`dv = dx => v = x`

`int ln(x^2+10x+24)dx = xln(x^2+10x+24) - int (2x^2 + 10x)/(x^2+10x+24) dx`

Using reminder theorem yields:

`(2x^2 + 10x) = 2(x^2+10x+24) - 10x`

`(2x^2 + 10x)/(x^2+10x+24) = 2 - 10x/(x^2+10x+24)`

Integrating both sides yields:

`int (2x^2 + 10x)/(x^2+10x+24) dx= int2 dx- int 10x/(x^2+10x+24) dx`

`int (2x^2 + 10x)/(x^2+10x+24) dx = 2x - 10 int x/(x^2+10x+24) dx`

You need to come up with the substitution `x^2+10x+24 = ` t such that:

`x^2+10x+24 = t => (2x + 10)dx = dt => (x + 5)dx = (dt)/2`

You need to add and subtract 5 to numerator such that:

`int x/(x^2+10x+24) dx = int (x + 5 - 5)/(x^2+10x+24) dx`

Using the linearity property yields:

`int x/(x^2+10x+24) dx = int (x + 5)/(x^2+10x+24) dx -5int 1/(x^2+10x+24) dx`

`int x/(x^2+10x+24) dx = int (dt)/(2t) - 5int 1/(x^2+10x+24) dx`

You need to convert the denominator `x^2+10x+24` to the form `x^2 +- a^2` .

You should complete the square `x^2+10x ` using the formula `(a+b)^2 =a^2 + 2ab + b^2` such that:

`x^2+10x = a^2 + 2ab => a = x; 2ab = 10x => 2xb = 10x => b = 5`

`x^2+10x+25 - 25 = (x+5)^2 - 25 `

`x^2+10x+24 = (x+5)^2 - 25 + 24 => x^2+10x+24 = (x+5)^2 - 1`

`int 1/(x^2 + a^2) dx = int 1/((x+5)^2 - 1)dx= int 1/(x+6)(x+4) dx`

`int 1/((x+5)^2 - 1) dx = ln|(x+4)/(x+6)| + c`

`int x/(x^2+10x+24) dx = (1/2)ln|x^2+10x+24| - 5ln|(x+4)/(x+6)| + c`

`int ln(x^2+10x+24)dx = xln(x^2+10x+24) - 2x + 10((1/2)ln|x^2+10x+24| - 5ln|(x+4)/(x+6)|) + c`

`int ln(x^2+10x+24)dx = xln(x^2+10x+24) - 2x + 5ln|x^2+10x+24| - 50ln|(x+4)/(x+6)| + c`

**Hence, evaluating the given indefinite integral yields** `int ln(x^2+10x+24)dx = xln(x^2+10x+24) - 2x + 5ln|x^2+10x+24| - 50ln|(x+4)/(x+6)| + c.`