review the 2nd Fundamental Theorem of Calculus.
If ∫ln(2t+1) dt = f(t), then
G(x) = f(x) - f(1)
So, naturally, G'(x) = ln(2x+1)
the 2nd integral is just
Given G of x equals the integral from 1 to x of the natural logarithm of the quantity 2 times t plus 1, dt, find G '(x)
ln(2x + 1)
ln(2x + 1) − ln3
ln(2t + 1) + C
2 divided by the quantity 2 times x plus 1
my answer is ln(2x+1) but I dont think its right
1 answer