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

Question

4463 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-10T13:07:03

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

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 · 2018-06-10T14:36:40

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

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.!