Evaluate the integral int(x+3)/(4x+4)dx. Choose 1 answer: (A) (1)/(4)x+ln |x+1|+C (B) (1)/(4)x+2ln |x+1|+C (C) (1)/(4)x+(ln |x+1|)/(2)+C (D) (1)/(4)x+(ln |x+1|)/(4)+C

Split Integral: Now, split the integral into two parts. 1/4 int(x/(x+1) + 3/(x+1))dx = 1/4 (int(x/(x+1)dx + int(3/(x+1)dx))Substitution and Integration: For the first part, int(x/(x+1)dx, we can use a simple substitution. Let u = x + 1, then du = dx and x = u - 1. Substitute into the integral. 1/4 int((u-1)/u du) = 1/4 (int(1 du) - int(1/u du))Final Integration: Now, integrate both parts. 1/4 (int(1 du) - int(1/u du)) = 1/4 (u - ln|u|) + CSubstitute Back: Substitute back for u = x + 1. 1/4 (u - ln|u|) + C = 1/4 (x + 1 - ln|x+1|) + C = 1/4 x + 1/4 - 1/4 ln|x+1| + CCombine Constants: Combine the constants into a single constant C. 1/4 x + 1/4 - 1/4 ln|x+1| + C = 1/4 x + (1/4 - 1/4) ln|x+1| + C = 1/4 x + ln|x+1|/4 + C

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Evaluate the integral
int(x+3)/(4x+4)dx.
Choose 1 answer:
(A)
(1)/(4)x+ln |x+1|+C
(B)
(1)/(4)x+2ln |x+1|+C
(C)
(1)/(4)x+(ln |x+1|)/(2)+C
(D)
(1)/(4)x+(ln |x+1|)/(4)+C

Evaluate the integral $\int \frac{x+3}{4 x+4} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{4} x+\ln |x+1|+C$ $\newline$ (B) $\frac{1}{4} x+2 \ln |x+1|+C$ $\newline$ (C) $\frac{1}{4} x+\frac{\ln |x+1|}{2}+C$ $\newline$ (D) $\frac{1}{4} x+\frac{\ln |x+1|}{4}+C$

View step-by-step help

Home

Math Problems

Calculus

Find indefinite integrals using the substitution

Full solution

Q. Evaluate the integral

\int \frac{x+3}{4 x+4} d x

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{4} x+\ln |x+1|+C

\newline

(B)

\frac{1}{4} x+2 \ln |x+1|+C

\newline

(C)

\frac{1}{4} x+\frac{\ln |x+1|}{2}+C

\newline

(D)

\frac{1}{4} x+\frac{\ln |x+1|}{4}+C

Split Integral: Now, split the integral into two parts. $\newline$ $\frac{1}{4} \int\left(\frac{x}{x+1} + \frac{3}{x+1}\right)dx = \frac{1}{4} \left(\int\frac{x}{x+1}dx + \int\frac{3}{x+1}dx\right)$
Substitution and Integration: For the first part, $\int\frac{x}{x+1}\,dx$ , we can use a simple substitution. Let $u = x + 1$ , then $du = dx$ and $x = u - 1$ . Substitute into the integral. $\frac{1}{4} \int\frac{(u-1)}{u} \,du$ = $\frac{1}{4} (\int 1 \,du - \int\frac{1}{u} \,du)$
Final Integration: Now, integrate both parts. $\newline$ $\frac{1}{4} (\int 1 \, du - \int \frac{1}{u} \, du) = \frac{1}{4} (u - \ln|u|) + C$
Substitute Back: Substitute back for $u = x + 1$ . $\newline$ $\frac{1}{4} (u - \ln|u|) + C = \frac{1}{4} (x + 1 - \ln|x+1|) + C$ $\newline$ = \frac{$1$}{$4$} x + \frac{$1$}{$4$} - \frac{$1$}{$4$} \ln|x+\(1| + C
Combine Constants: Combine the constants into a single constant $C$ . $\frac{1}{4} x + \frac{1}{4} - \frac{1}{4} \ln|x+1| + C = \frac{1}{4} x + (\frac{1}{4} - \frac{1}{4}) \ln|x+1| + C = \frac{1}{4} x + \frac{\ln|x+1|}{4} + C$