Use the table of integrals in this section to find the indefinite integral. ( I keep getting this answer wrong and I don't know why)

Question

Answer 1

the answer is (3 √(1 + (x-4)²)/(x-4 )+C

1/[(x-4)²√(x²-8x+17)]=1/[(x-4)²√(1 + (x-4)²)

Now if you substitute x-4=tant, then dx=dt/cos²(x) and

∫3/[(x-4)²√(1 + (x-4)^√2)dx=3∫cod(t)/(sin²(t)) dt=-3/sin(t) = |

Since tanx=y, hence sint=(x-4)/(√(1 + (x-4)²) |

=--3√(1 + (x-4)²)/(x-4)+C