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

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### 1 Answer

You should 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 x(2x+10)/(x^2+10x+24) dx`

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

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

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

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

You should use partial fraction decomposition to split the fraction `(5x + 24)/(x^2+10x+24)` into more simpler fractions such that:

`(5x + 24)/(x^2+10x+24) = (5x + 24)/((x + 4)(x + 6))`

`(5x + 24)/((x + 4)(x + 6)) = A/(x+4) + B/(x+6)`

Bringing the terms to a common denominator yields:

`5x + 24 = A(x + 6) + B(x + 4)`

`5x + 24 = Ax + 6A + Bx + 4B`

`5x + 24 = x(A + B) + 6A + 4B`

Equating the coefficients of like powers yields:

`A+B = 5`

`6A + 4B = 24 => 3A + 2B = 12`

You need to multiply the first equation by -2 such that:

`-2A - 2B = -10`

You need to add this equation to `3A + 2B = 12` such that:

`3A + 2B - 2A - 2B = 12 - 10`

`A = 2 => B = 5-2 => B = 3`

`(5x + 24)/((x + 4)(x + 6)) = 2/(x+4) + 3/(x+6)`

Integrating both sides yields:

`int (5x + 24)/((x + 4)(x + 6)) dx= 2int 1/(x+4) dx+ 3int 1/(x+6) dx`

`int (5x + 24)/((x + 4)(x + 6)) dx = 2ln|x+4| + 3ln|x+6| + c`

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

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

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