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Chapter 1: Problem 1

Evaluate the following integrals: (i) \(\int \frac{\mathrm{dx}}{(2 \mathrm{x}+1) \sqrt{(4 \mathrm{x}+3)}}\) (ii) \(\int \frac{1}{(x-3) \sqrt{x+1}} \mathrm{dx}\)

Short Answer

Expert verified
Question: Evaluate the following integrals: (i) $$\int \frac{\mathrm{dx}}{(2 \mathrm{x}+1) \sqrt{(4 \mathrm{x}+3)}}$$ (ii) $$\int \frac{1}{(x-3) \sqrt{x+1}} \mathrm{dx}$$ Answer: (i) $$\int \frac{\mathrm{dx}}{(2 \mathrm{x}+1) \sqrt{(4 \mathrm{x}+3)}} = \sqrt{4x+3} - \ln|\sqrt{4x+3}+1| + C$$ (ii) $$\int \frac{1}{(x-3) \sqrt{x+1}} \mathrm{dx} = 2\sqrt{x+1}+2\ln|\sqrt{x+1}-2| -2\ln|x-3| + C$$

Step by step solution

01

(i) Integral 1: Choose the substitution u

Let \(u = 4x + 3\). This will simplify the expression under the square root.
02

(i) Differentiate, and solve for dx

Differentiate u with respect to x: \(du/dx = 4\). Then solve for dx: \(dx = \frac{1}{4}du\).
03

(i) Replace dx and rearrange

In the integral, substitute u in the expression and replace dx with the expression in terms of u: $$\int \frac{\mathrm{dx}}{(2 \mathrm{x}+1) \sqrt{(4 \mathrm{x}+3)}} = \int \frac{1}{(2x + 1) \sqrt u} \cdot \frac{1}{4} du$$
04

(i) Simplify and apply the substitution

The original substitution was \(u = 4x + 3 = 2(2x+1) - 1\). Therefore, \(2x+1=\frac{u+1}{2}\).$$\int \frac{1}{(2x + 1) \sqrt u} \cdot \frac{1}{4} du = \int \frac{1}{(\frac{u+1}{2}) \sqrt u} \cdot \frac{1}{4} du$$
05

(i) Simplify and separate terms

$$\int \frac{2}{u+1} \cdot \frac{1}{\sqrt u} \cdot \frac{1}{4} du= \int \frac{1}{2(u+1)\sqrt u} du$$
06

(i) Evaluate the integral

The integral can now be expressed in terms of a standard form:$$\int \frac{1}{2(u+1)\sqrt u} du = \frac{1}{2} \int \frac{du}{(u+1)\sqrt u}$$Now, you may use rational function integration techniques such as partial fraction decomposition or lookup a table of integrals to find the antiderivative:$$\frac{1}{2} \left(\int \frac{du}{(u+1)\sqrt u}\right) = \frac{1}{2} (2\sqrt{u}-2\ln|\sqrt{u}+1| + C)=\sqrt{u} - \ln|\sqrt{u}+1| + C$$
07

(i) Back-substitute

Now back-substitute \(u = 4x + 3\):$$\sqrt{4x+3} - \ln|\sqrt{4x+3}+1| + C$$
08

(ii) Integral 2: Choose the substitution v

Let \(v = x + 1\). This will simplify the expression under the square root.
09

(ii) Differentiate, and solve for dx

Differentiate v with respect to x: \(dv/dx = 1\). Then solve for dx: \(dx = dv\).
10

(ii) Replace dx and rearrange

In the integral, substitute v in the expression and replace dx with the expression in terms of v: $$\int \frac{1}{(x-3) \sqrt{x+1}} \mathrm{dx} = \int \frac{1}{(x-3) \sqrt v} dv$$
11

(ii) Apply the substitution

Since \(v = x + 1\), we have \(x = v - 1\). Substitute this expression for x: $$\int \frac{1}{(x-3) \sqrt v} dv = \int \frac{1}{(v-4)\sqrt v} dv$$
12

(ii) Evaluate the integral

The integral can now be expressed in terms of a standard form: $$\int \frac{1}{(v-4)\sqrt v} dv$$Now, you may use rational function integration techniques such as partial fraction decomposition or lookup a table of integrals to find the antiderivative:$$2\sqrt{v}+2\ln|\sqrt{v}-2| -2\ln|v-4| + C$$
13

(ii) Back-substitute

Now back-substitute \(v = x + 1\):$$2\sqrt{x+1}+2\ln|\sqrt{x+1}-2| -2\ln|x-3| + C$$

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