f(x)= 7/x^3-9/x^5. let F(x) be antiderivative of f(x) with F(1)=0 I got a c=1.25 but is wrong. need help, please

Question

f(x)= 7/x^3-9/x^5. let F(x) be antiderivative of f(x) with F(1)=0    I got a c=1.25 but is wrong. need help, please

anonymous · Answer

I get F(x)= 9/4x^4-7/2x^2 +c

anonymous · Answer

For F(1) = 2.25-3.5+c=0. so C=1.25,right?

anonymous · Answer

i didnt reduce my stupid fraction

anonymous · Answer

of all the petty things i screw up! How do i give medals?

anonymous · Answer

where did ur reply go?

anonymous · Answer

$$F(x) = \frac{-7}{2x^2}+\frac{9}{4x^4}+c$$

You were correct

Then$$F(1) = c - \frac{5}{4}$$

So c = 1.25

Nothing is wrong here

anonymous · Answer

So why did you think your answer is wrong?

anonymous · Answer

it doesnt check with webwork

anonymous · Answer

I believe you were correct, unless f(x) is not as same as you wrote :)

anonymous · Answer

its right. Thanks for the reassurance!

anonymous · Answer

How do i give medals to my tutors, like you?

anonymous · Answer

Press "Best Response" if you think my answer is deserved. Actually, I remove your doubts due to I love maths only, not because of loving medal