Question

Part B: 1. Use partial fractions to evaluate the integral $\int \frac{1}{(x+4)(x-1)} dx$ 2. Use partial fractions to evaluate the integral $\int_2^3 \frac{1}{x^2 - 1} dx = \int_2^3 \frac{1}{(x-1)(x+1)} dx$ 3. Use partial fractions to evaluate the integral $\int \frac{x+2}{(x+5)(x-1)} dx$

Name: part b 1 use partial fractions to evaluate the integral int frac1x4x 1 dx 2 use partial fractions to evaluate the integral int23 frac1x2 1 dx int23 frac1x 1x1 dx 3 use partial fractions to 83772
Uploaded: 2024-10-08T09:03:16-08:00
Duration: 4 min 12 s
Channel: Shu Naito
Description: part b 1 use partial fractions to evaluate the integral int frac1x4x 1 dx 2 use partial fractions to evaluate the integral int23 frac1x2 1 dx int23 frac1x 1x1 dx 3 use partial fractions to 83772

          Part B:
1. Use partial fractions to evaluate the integral $\int \frac{1}{(x+4)(x-1)} dx$
2. Use partial fractions to evaluate the integral $\int_2^3 \frac{1}{x^2 - 1} dx = \int_2^3 \frac{1}{(x-1)(x+1)} dx$
3. Use partial fractions to evaluate the integral $\int \frac{x+2}{(x+5)(x-1)} dx$

$part b 1 use partial fractions to evaluate the integral int frac1x4x 1 dx 2 use partial fractions to evaluate the integral int23 frac1x2 1 dx int23 frac1x 1x1 dx 3 use partial fractions to 83772$

Added by Sarah C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Shu Naito

10/08/2024

Step 1

We have $\frac{1}{(x+4)(x-1)} = \frac{A}{x+4} + \frac{B}{x-1}$ Multiplying both sides by $(x+4)(x-1)$, we get $1 = A(x-1) + B(x+4)$ If $x=1$, then $1 = 5B$, so $B = \frac{1}{5}$. If $x=-4$, then $1 = -5A$, so $A = -\frac{1}{5}$. Therefore, $\frac{1}{(x+4)(x-1)} = Show more…

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Part B: 1. Use partial fractions to evaluate the integral $\int \frac{1}{(x+4)(x-1)} dx$ 2. Use partial fractions to evaluate the integral $\int_2^3 \frac{1}{x^2 - 1} dx = \int_2^3 \frac{1}{(x-1)(x+1)} dx$ 3. Use partial fractions to evaluate the integral $\int \frac{x+2}{(x+5)(x-1)} dx$