If I see the derivative of the base sitting out front either directly of as a multiple, I usually can do the integral intuitively
I know it has to be (?)(1 + x^2)^(3/2)
I then visulaize differentiating that, and adjust the coefficient accordingly.
so it should be (1/3)(1 + x^2)^(3/2)
Your teacher probably wants you to do it by substitution
let u = (1 + x^2)
du/dx = 2x
dx = du/2x
so Integral of x sq/(1+x sq) dx
= integral (1/2)u^(1/2) du
= (1/3)u^(3/2)
= (1/3)(1 + x^2)^(3/2)
Integral of x sq/(1+x sq) dx
I am not sure how to do this, am i supposed to do partial fractions to break it up? Or is it just some simple substitution?
2 answers
But how did you get
Integral of x sq/(1+x sq) dx
= integral (1/2)u^(1/2) du
du= 2xdx, from there i got,
integral xsq/(1+x) dx
= (1/2) integral[ (sqrt (1-u))/u ]du
Did i do anything wrong?
Integral of x sq/(1+x sq) dx
= integral (1/2)u^(1/2) du
du= 2xdx, from there i got,
integral xsq/(1+x) dx
= (1/2) integral[ (sqrt (1-u))/u ]du
Did i do anything wrong?
1 answer