int sqrt(x^2+5/x)=∫√x2+5x=?

1 Answer
Ananda Dasgupta
Mar 31, 2018

sqrt(x^2+5) +5/(2sqrt5) ln((sqrt(x^2+5)-sqrt5)/(sqrt(x^2+5)+sqrt5))+C√x2+5+52√5ln(√x2+5−√5√x2+5+√5)+C

Explanation:

Substitute x^2+5=u^2x2+5=u2. Then, 2xdx=2udu2xdx=2udu and thus dx/x =(udu)/x^2=(udu)/(u^2-5)dxx=udux2=uduu2−5.

So, our integral becomes

int sqrt(x^2+5)/x dx = int u (udu)/(u^2-5)∫√x2+5xdx=∫uuduu2−5
qquad = int (u^2-5+5)/(u^2-5) = int (1+5/(u^2-5))du
qquad = u+5 int (du)/(u^2-(sqrt5)^2)= u + 5/(2sqrt5) ln((u-sqrt5)/(u+sqrt5))
qquad = sqrt(x^2+5) +5/(2sqrt5) ln((sqrt(x^2+5)-sqrt5)/(sqrt(x^2+5)+sqrt5))

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