Question

Integral X(arctanx)^2 dx

Answer 1

∫ sqrt(a^2 + x^2) dx

= ∫ sqrt(a^2 + (au)^2) * a * du

= ∫ a * sqrt(1 + u^2) * a * du

= (a^2) * ∫ sqrt(1 + u^2) du

= 0.5*a^2*[ u*sqrt(1 + u^2) + Ln(u + sqrt(1 + u^2)) ] + c

Now making the reverse substitution u = x/a, we get

0.5*a*x*sqrt(1 + (x/a)^2) + 0.5*a^2*Ln(x/a + sqrt(1 + (x/a)^2)) + c

With a little algebra you can express this in an alternate form that is more compact:

[x*sqrt(x^2 + a^2) + a^2*Ln(x + sqrt(x^2 + a^2))]/2 + c

Answer 2

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Hey

See the attachment
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