Question

How do you integrate `int (x^2 +x +5)/(sqrt(x^2 +1)) dx`?

Answer 1

`1/4t^2-t+4ln(t^2+1)+2arctan(t)+1/2*ln(t)+C`
where `t=sqrt(x^2+1)-x`

Explanation:

Substituting

`sqrt(x^2+1)=t+x`
then

`x=(1-t^2)/(2t)`
and
`dx=-1/2*(t^2+1)/(2t^2)dt`

so we get the integral

`int(1/2t-1+1/(2t)+(8t+2)/(t^2+1))dt`
which gives

`+1/4t^2-t+4ln(t^2+1)+2arctan(t)+1/2*ln(t)+C`
where
`t=sqrt(x^2+1)-x`

Answer 2

`1/2[(x+2)sqrt(x^2+1)+9ln|(x+sqrt(x^2+1))|]+C`.

Explanation:

Let, `I=int(x^2+x+5)/sqrt(x^2+1)dx=int{(x^2+1)+(x+4)}/sqrt(x^2+1)dx`.

`:. I=int{(x^2+1)/sqrt(x^2+1)+(x+4)/sqrt(x^2+1)}dx`,

`=intsqrt(x^2+1)dx+int(x+4)/sqrt(x^2+1)dx`,

`=I_1+1/2int(2x)/sqrt(x^2+1)dx+4int1/sqrt(x^2+1)dx`,

`:. I=I_1+1/2I_2+4I_3..........................(star)`, where,

`I_1=intsqrt(x^2+1)dx`,

`:. I_1=x/2sqrt(x^2+1)+1/2ln|(x+sqrt(x^2+1))|...........(star_1)`;

`I_2=int(2x)/sqrt(x^2+1)dx`,

`=int(x^2+1)^(-1/2)d/dx(x^2+1)dx`,

`=(x^2+1)^(-1/2+1)/(-1/2+1)`,

`:. I_2=2sqrt(x^2+1)....................................(star_2)`;

`I_3=int1/sqrt(x^2+1)dx`,

`:. I_3=ln|x+sqrt(x^2+1)|..............................(star_3)`.

`"Using "(star_1), (star_2)" and "(star_3)" in "(star)," we have,"`

`I=x/2sqrt(x^2+1)+1/2ln|(x+sqrt(x^2+1))|+1/2*2sqrt(x^2+1)`

`+4*ln|x+sqrt(x^2+1)|, i.e.,`

`I=1/2[(x+2)sqrt(x^2+1)+9ln|(x+sqrt(x^2+1))|]+C`.

Enjoy Maths.!