find intergral (x^2)/sqrt(x^2+25)
I gave up working on it
Wolfram gave me this
http://integrals.wolfram.com/index.jsp?expr=%28x%5E2%29%2Fsqrt%28x%5E2%2B25%29&random=false
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inegrate (x^2)/sqrt(x^2+25)
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To find the integral of (x^2)/sqrt(x^2+25), we can start by using a u-substitution.
Let u = x^2 + 25.
Then, du = 2x dx, and dx = du / (2x).
Substituting these values into the integral, we have:
∫ (x^2)/sqrt(x^2+25) dx
= ∫ (x^2)/sqrt(u) * (du / (2x))
= (1/2) ∫ (x^2/sqrt(u)) du.
We can simplify the expression further by canceling out one of the x terms:
= (1/2) ∫ (x/sqrt(u)) * x du
= (1/2) ∫ sqrt(x) * sqrt(x) / sqrt(u) du
= (1/2) ∫ sqrt(x^2) / sqrt(u) du
= (1/2) ∫ 1 / sqrt(u) du.
Integrating 1 / sqrt(u) with respect to u gives us:
= (1/2) * 2√u + C
= √u + C.
Now, we need to substitute back for u:
= √(x^2 + 25) + C.
Therefore, the integral of (x^2)/sqrt(x^2+25) is √(x^2 + 25) + C, where C is the constant of integration.