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

2 Answers
Jun 10, 2018

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

Jun 10, 2018

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.!