Evaluate the integral int(x-5)/(-2x+2)dx. Choose 1 answer: (A) (1)/(2)x-2ln |x-1|+C (B) (1)/(2)x+2ln |x-1|+C (C) -(1)/(2)x-2ln |x-1|+C (D) -(1)/(2)x+2ln |x-1|+C

Rewrite Integral: Rewrite the integral as the sum of two simpler integrals. int(x-5)/(-2x+2)dx = int(1/2 - 5/(-2x+2))dxSplit into Parts: Split the integral into two parts. int(1/2 - 5/(-2x+2))dx = int(1/2)dx - int(5/(-2x+2))dxIntegrate First Part: Integrate the first part. int(1/2)dx = (1/2)xMake Substitution: Make a substitution for the second part, let u = -2x + 2. du = -2dx, so dx = -1/2 duChange Limits: Change the limits of integration according to the substitution. int(5/(-2x+2))dx = -5/2 int(1/u)duIntegrate Second Part: Integrate the second part using the substitution. -5/2 int(1/u)du = -5/2 ln|u| + CSubstitute Back: Substitute back for u. -5/2 ln|u| + C = -5/2 ln|-2x+2| + CCombine Integrals: Combine the two parts of the integral. (1/2)x - 5/2 ln|-2x+2| + CSimplify Expression: Simplify the expression. (1/2)x - 5/2 ln|-2x+2| + C = (1/2)x + (5/2)ln|x-1| + C

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Evaluate the integral
int(x-5)/(-2x+2)dx.
Choose 1 answer:
(A)
(1)/(2)x-2ln |x-1|+C
(B)
(1)/(2)x+2ln |x-1|+C
(C)
-(1)/(2)x-2ln |x-1|+C
(D)
-(1)/(2)x+2ln |x-1|+C

Evaluate the integral $\int \frac{x-5}{-2 x+2} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2} x-2 \ln |x-1|+C$ $\newline$ (B) $\frac{1}{2} x+2 \ln |x-1|+C$ $\newline$ (C) $-\frac{1}{2} x-2 \ln |x-1|+C$ $\newline$ (D) $-\frac{1}{2} x+2 \ln |x-1|+C$

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Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Evaluate the integral

\int \frac{x-5}{-2 x+2} d x

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{2} x-2 \ln |x-1|+C

\newline

(B)

\frac{1}{2} x+2 \ln |x-1|+C

\newline

(C)

-\frac{1}{2} x-2 \ln |x-1|+C

\newline

(D)

-\frac{1}{2} x+2 \ln |x-1|+C

Rewrite Integral: Rewrite the integral as the sum of two simpler integrals. $\int(\frac{x-5}{-2x+2})dx = \int(\frac{1}{2} - \frac{5}{-2x+2})dx$
Split into Parts: Split the integral into two parts. $\int(\frac{1}{2} - \frac{5}{-2x+2})dx = \int(\frac{1}{2})dx - \int(\frac{5}{-2x+2})dx$
Integrate First Part: Integrate the first part. $\int \frac{1}{2}dx = \frac{1}{2}x$
Make Substitution: Make a substitution for the second part, let $u = -2x + 2$ . $du = -2dx$ , so $dx = -\frac{1}{2} du$
Change Limits: Change the limits of integration according to the substitution. $\newline$ $\int\frac{5}{(-2x+2)}\,dx = -\frac{5}{2} \int\frac{1}{u}\,du$
Integrate Second Part: Integrate the second part using the substitution. $\newline$ -\frac{$5$}{$2$} \int(\frac{$1$}{u})du = -\frac{$5$}{$2$} \ln|u| + C
Substitute Back: Substitute back for \(u. $-\frac{5}{2} \ln|u| + C = -\frac{5}{2} \ln|-2x+2| + C$
Combine Integrals: Combine the two parts of the integral. $(\frac{1}{2})x - \frac{5}{2} \ln|{-2x+2}| + C$
Simplify Expression: Simplify the expression. $\newline$ (\frac{$1$}{$2$})x - \frac{$5$}{$2$} \ln|{$-2$x+$2$}| + C = (\frac{$1$}{$2$})x + (\frac{$5$}{$2$})\ln|x\(-1| + C