Question

Evaluate the integral I = ? (3x + 3) / ((x + 2)(x - 1)) dx. A. I = -ln|x + 2| + 2ln|x - 1| + c, where c is a constant. B. I = -ln|x + 2| - 2ln|x - 1| + c, where c is a constant. C. I = ln|x + 3| + 2ln|x - 2| + c, where c is a constant. D. I = ln|x + 2| - 2ln|x - 1| + c, where c is a constant. E. I = ln|x + 2| + 2ln|x - 1| + c, where c is a constant.

          Evaluate the integral I = ? (3x + 3) / ((x + 2)(x - 1)) dx.
A. I = -ln|x + 2| + 2ln|x - 1| + c, where c is a constant.
B. I = -ln|x + 2| - 2ln|x - 1| + c, where c is a constant.
C. I = ln|x + 3| + 2ln|x - 2| + c, where c is a constant.
D. I = ln|x + 2| - 2ln|x - 1| + c, where c is a constant.
E. I = ln|x + 2| + 2ln|x - 1| + c, where c is a constant.

Evaluate the integral I = ? (3x + 3) / ((x + 2)(x - 1)) dx.
A. I = -ln|x + 2| + 2ln|x - 1| + c, where c is a constant.
B. I = -ln|x + 2| - 2ln|x - 1| + c, where c is a constant.
C. I = ln|x + 3| + 2ln|x - 2| + c, where c is a constant.
D. I = ln|x + 2| - 2ln|x - 1| + c, where c is a constant.
E. I = ln|x + 2| + 2ln|x - 1| + c, where c is a constant.

Added by Bethany R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert J Hardin

02/13/2018

Step 1

Evaluate the integral / = 3x+3 Sr+zx 11) dx OA / = Inlx+2/+2In/x-1/+c,where € is a constant OB./= - Inlx+2| 2Inlx-1/+ c, Where € is a constant. Oc /= Inlx+3l+2Inlx-2/+ €, where € is a constant: OD /= Inlx+2/ - ZInlx-1/+c, where € is a constant 9E /=Inlx+2/ Show more…

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Evaluate the integral I = ∫ (3x + 3) / ((x + 2)(x - 1)) dx. A. I = -ln|x + 2| + 2ln|x - 1| + c, where c is a constant. B. I = -ln|x + 2| - 2ln|x - 1| + c, where c is a constant. C. I = ln|x + 3| + 2ln|x - 2| + c, where c is a constant. D. I = ln|x + 2| - 2ln|x - 1| + c, where c is a constant. E. I = ln|x + 2| + 2ln|x - 1| + c, where c is a constant.