Question

Integration of X+√x+1/x+2 dx

Answer 1

hi,

Let I=∫x+x+1√x+2dx

=∫xx+2dx+∫x+1√x+2dx

=I1+I2

Where I1=∫xx+2dx

and I2=∫x+1√x+2dx

Lets solve I1 first

So, I1=∫x+2−2x+2dx

=∫dx−∫2x+2dx

=x−2ln(x+2)

Now, lets solve I2,

I2=∫x+2−1√x+2dx

Let x+2=sec2(y)

⟹dx=2sec2(y)tan(y)dy

Now substituting x, dx, we have

I2=∫2sec2(y)−1√sec2(y)tan(y)sec2(y)dy

=∫2sec2(y)tan2(y)sec2(y)dy (As sec2(y)−1=tan2(y))

=∫2tan2(y)dy

=∫2(sec2(y)−1)dy

=∫2sec2(y)dy−∫2dy

=2tan(y)−2y

As, x+2=sec2(y)

⟹x+1=sec2(y)−1

⟹x+1=tan2(y)

⟹x+1−−−−√=tan(y)

y=arctanx+1−−−−√

So substituting the value of y into I2, we have,

I2=2tan(arctanx+1−−−−√)−2arctanx+1−−−−√

=2x+1−−−−√−2arctanx+1−−−−√

So, I=x−2ln(x+2)+2x+1−−−−√−2arctanx+1−−−−√