Evaluate the indefinite integral.

Question

Evaluate the indefinite integral.

∫6 / (t+3)^5 dt

 + C

Having trouble learning this one

Answer 1

Hello Jay,

You can re-write that integral as ∫6 *(t+3)^-5 dt

let u = t+3

then du = 1dt = dt

So you can substitute the integral as ∫6 *u^-5 du

∫6 *u^-5 du = 6∫u^-5 du

Use the power rule of integration: 6∫u^-5 du = 6* u^-5+1/(-5+1) + C= 6 u^-4/-4 + C = (6/-4)u^-4 + C = (-3/2)*u^-4+C

Now substitute the original function for u: (-3/2)*u^-4+C= (-3/2)*(t+3)^-4 + C

which you can also write as -3/(2*(t+3)⁴) + C

Conclusion: ∫6 / (t+3)^5 dt = (-3/2)*(t+3)^-4 + C

alternate way to write the answer: ∫6 / (t+3)^5 dt = -3/(2*(t+3)⁴) + C