How do you evaluate #int# #arctan(sqrt(x))/sqrt(x)# dx?

Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)

#int# #arctan(sqrt(x))/sqrt(x)# dx

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1 Answer

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Explanation:

Use the u substitution.

u = #sqrt(x)#

du = #1/(2sqrt(x))# dx

2du = #1/sqrt(x)# dx

Write the new formula after the u substitution.

2 #int# #tan^-1(u)# du

Use table 89 to find the integral of 2#tan^-1(u)#.

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2 #int# #tan^-1(u)# du
= 2[u #tan^-1(u)# - #1/2# ln(1 + #u^2#)] + C

Replace the u variable back in the terms of x.

= 2[#sqrt(x)# #tan^-1(sqrt(x))# - #1/2# ln(1 + #sqrt(x)^2#)] + C

Simplify the answer.

= 2[#sqrt(x)# #tan^-1(sqrt(x))# - #1/2# ln(1 + #x#)] + C

= 2#sqrt(x)# #tan^-1(sqrt(x))# - ln(1 + #x#) + C