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(1+x^2)^(5/2)
(1+cos(x))
(1+x^2)
(x^3+x), partial fraction or MAYBE NOT??
(x+1)^(3/2) from 0 to inf
(1+sqrt(x)) (substitution)
(ax^2+bx+c) (ALL CASES 💪)
Differentiate in both sides
sqrt(x^2+16)
sqrt(ax+b) (substitution)
Apply 1 = sin^2(u) + cos2(u)
Apply sin(2u) = 2sin(u)cos(u)
(x+1)^2 (substitution)
(x^2+1) from -inf to inf, Contour Integral
(a^2+x^2)
(1+x^2)^(3/2) (substitution)
(x^2+1) but you didn't learn it this way in calculus 2
(x^3+1) from 100 integrals
sqrt (x^2 + a^2)
(x^(1/2)+x^(1/4)) (substitution)
(ln(x))^2
(x^3+x), partial fraction or MAYBE NOT
Substitute x and dx
sqrt(x)
(a^2+x^2)^(3/2) (substitution)
tan(x) (substitution)
x from 1 to Infinity
(cos^2(x) + 2sin^2(x)) dx
Rewrite expression
Apply sin(arctan(x)) and cos(arctan(x)) formulas
(x^6+1) without partial fractions!
(16+9x^2) (substitution)
sec(x)
x(1 + ln x)
(x^3+x)
(9+x^2)^(3/2)
(4+x^2)
Trig Substitution
(1+tan(x)) (substitution)
(x*(ln(x))^3) (substitution)
(x*(1+(ln(x))^2)) (substitution)
(1+sqrt(2x))
(x*sqrt(ln(x)) from 1 to e
(4 + x^2)
(1 + cos x) dx
(1+sqrt(x))
(x^(1/2) + x^(1/3)) (substitution)
(x^2+9) (substitution)
(x*ln(x)) (substitution)
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