Question

Integration of x/(2x+5)?

Answer 1

`I=1/4((2x+5)-5ln(abs((2x+5))))+C`

Explanation:

We want to solve

`I=intx/(2x+5)dx`

Make a substitution `color(blue)(u=2x+5=>du=2dx`

`I=1/2intx/udu`

But `color(blue)(u=2x+5=>x=(u-5)/2`

`I=1/2int((u-5)/2)/udu`

`color(white)(I)=1/4int(u-5)/udu`

`color(white)(I)=1/4int1-5/udu`

`color(white)(I)=1/4(u-5ln(abs(u)))+C`

Substitute back `u=(2x+5)`

`I=1/4((2x+5)-5ln(abs((2x+5))))+C`

Answer 2

`I=intx/(2x+5)dx`

Let `2x+5 = t`

Therefore, `dt = 2 dx => dx=dt/2`

Also, `x=(t-5)/2`

`=>I=int(t-5)/(2(t)) dt/2`

`=>1/4int(t-5)/t dt`

`=>1/4[int1dt-5int1/t dt]`

`=>1/4[t-5log|t|] +c`

`=>1/4(2x+5)-5/4log|2x+5 | +c`