Question #13763

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Answer 1 · 2017-06-18T15:02:52

#1/atan^-1(x/a) + "constant"#

Factor out an #a^2#

#int 1/(x^2+a^2)dx=int 1/(a^2((x^2)/(a^2)+1))dx#

Factor out a constant

#=1/a^2int1/((x^2)/(a^2)+1)dx#

Let #u=x/a# and #du=1/a dx#, so that #adu=dx#

#=1/a^cancel(2)intcancel(a)/(u^2+1)du#

#=1/aint1/(u^2+1)du#

#=1/a tan^-1(u)+"constant"#

#=1/atan^-1(x/a) + "constant"#

Answer 2 · 2017-06-18T15:08:29

Substitute #x = atan(theta), dx = asec^2(theta)d theta#

Given: #int1/(x^2+a^2)dx =#

Substitute #x = atan(theta), dx = asec^2(theta)d theta#

#int(asec^2(theta))/((atan(theta))^2+a^2)d theta =#

#int(asec^2(theta))/(a^2tan^2(theta)+a^2)d theta =#

#int(sec^2(theta))/(a(tan^2(theta)+1))d theta =#

We know that #sec^2(theta)=(tan^2(theta)+1)# so the integrand becomes #1/a#:

#1/aintd theta = 1/atheta+C#

Solve the substitution for #theta# and then substitute:

#x = atan(theta)#

#x/a = tan(theta)#

#theta = tan^-1(x/a)#

#int1/(x^2+a^2)dx = 1/atan^-1(x/a)+C#