How do you evaluate sqrt(64 + x^2)/x?

Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)

int sqrt(64 + x^2)/x dx

1 Answer
Noah G
Jan 18, 2018

I = sqrt(x^2 +64) - 8ln|sqrt(x^2 + 64)/x + 8/x| + C

Explanation:

Use the trig substitution x = 8tantheta. This means that dx = 8sec^2theta d theta

I = int sqrt(64 + (8tantheta)^2)/(8tantheta) * 8sec^2theta d theta

I = int sqrt(64 + 64tan^2theta)/(8tantheta) * 8sec^2theta d theta

I = int sqrt(64(1 + tan^2theta))/(8tantheta) * 8sec^2theta d theta

I = int sqrt(64sec^2theta)/tantheta sec^2theta d theta

I = int (8sec^3theta)/tantheta d theta

I = int 8sec^3theta cot theta d theta

I = 8 int 1/costheta(sec^2theta)costheta/sintheta d theta

I = 8 int sec^2theta/sintheta d theta

I = 8int sec^2theta csctheta d theta

I = 8 int (1 + tan^2theta)csctheta d theta

I = 8int csctheta + sin^2theta/cos^2theta(1/sintheta) d theta

I = 8 int cscthetad theta + 8int sintheta/cos^2theta d theta

I = 8int csctheta d theta + 8 inttanthetasectheta d theta

These are two known integrals. We know that int cscx dx = -ln(cscx + cotx) and int secxtanxdx = secx. Thus:

I = 8sectheta - 8ln|csctheta + cottheta| + C

From our initial substitution, we know that tantheta = x/8. Therefore, if we were to draw a right triangle we would find the hypotenuse to be sqrt(x^2 + 8^2) = sqrt(x^2 + 64). Thus, sectheta = sqrt(x^2 + 64)/8, csctheta = sqrt(x^2+ 64)/x and cottheta = 8/x.

I = 8sqrt(x^2 + 64)/8 - 8ln|sqrt(x^2+ 64)/x + 8/x| + C

I = sqrt(x^2 +64) - 8ln|sqrt(x^2 + 64)/x + 8/x| + C

Hopefully this helps!

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