First, perform division to get:
(-4x^2 + 5)/(2x + 3) = (-2x + 3) - 4/(2x + 3).
Then, integrating term-by-term yields:
∫ (-4x^2 + 5)/(2x + 3) dx = ∫ [(-2x + 3) - 4/(2x + 3)] dx
= ∫ -2x dx + ∫ 3 dx - ∫ 4/(2x + 3) dx
= -2 ∫ x dx + 3 ∫ dx - 4 ∫ 1/(2x + 3) dx, by pulling out constants
= (-2)(1/2)x^2 + 3x - (4)(1/2)ln|2x + 3| + C, by integrating
= -x^2 + 3x - 2ln|2x + 3| + C.
Notice that ∫ 1/(2x + 3) dx can be computed using:
u = 2x + 3 ==> du = 2 dx,
which gives ∫ 1/(2x + 3) dx = (1/2)ln|2x + 3| + C.
The first and second integrals we computed using the Power Rule in reverse.
I hope this helps!
no sure about ans #1
the derivative of an integral of f[x] = f[x]
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First, perform division to get:
(-4x^2 + 5)/(2x + 3) = (-2x + 3) - 4/(2x + 3).
Then, integrating term-by-term yields:
∫ (-4x^2 + 5)/(2x + 3) dx = ∫ [(-2x + 3) - 4/(2x + 3)] dx
= ∫ -2x dx + ∫ 3 dx - ∫ 4/(2x + 3) dx
= -2 ∫ x dx + 3 ∫ dx - 4 ∫ 1/(2x + 3) dx, by pulling out constants
= (-2)(1/2)x^2 + 3x - (4)(1/2)ln|2x + 3| + C, by integrating
= -x^2 + 3x - 2ln|2x + 3| + C.
Notice that ∫ 1/(2x + 3) dx can be computed using:
u = 2x + 3 ==> du = 2 dx,
which gives ∫ 1/(2x + 3) dx = (1/2)ln|2x + 3| + C.
The first and second integrals we computed using the Power Rule in reverse.
I hope this helps!
no sure about ans #1
the derivative of an integral of f[x] = f[x]