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Chapter 1: Problem 8

Evaluate the following integrals: $$ \int \frac{(3 x+5) d x}{\sqrt{x^{2}+4 x+3}} $$

Short Answer

Expert verified
Question: Evaluate the integral of the given function: $$ \int \frac{(3x+5) dx}{\sqrt{x^2 + 4x + 3}} $$ Answer: $$ \int \frac{(3x+5) dx}{\sqrt{x^2 + 4x + 3}} = \sinh^{-1}(3x + 5) + C $$

Step by step solution

01

Complete the square in the denominator

First, let's complete the square in the denominator. We have: $$ x^2 + 4x + 3 = x^2 + 4x + 4 - 1 = (x+2)^2 - 1 $$ Now the integral becomes: $$ \int \frac{(3x+5) dx}{\sqrt{(x+2)^2 - 1}} $$
02

Apply the substitution method

Let \(u = x+2\). Then it implies \(du = dx\) and \(x=u-2\). Substitute these into the integral: $$ \int \frac{(3(u-2)+5) du}{\sqrt{u^2 - 1}} $$ Simplify the expression in the numerator: $$ \int \frac{(3u-1) du}{\sqrt{u^2 - 1}} $$
03

Perform integration

Now notice how the numerator is just the derivative of the denominator's inside function. So we can perform the integration by using the following \(arcsinh\) formula: $$ \int \frac{u'}{\sqrt{u^2 - a^2}} du = \sinh^{-1}\left(\frac{u}{a}\right) + C $$ In this case, we have \(u' = 3u - 1\), \(u = 3u-1\) and \(a = 1\). Applying the formula to our original integral, we get: $$ \int \frac{(3u-1) du}{\sqrt{u^2 - 1}} = \sinh^{-1}\left(\frac{3u-1}{1}\right) + C $$
04

Replace the substitution variable

Now we need to replace the substitution variable \(u\) with our original variable \(x\). Remember that we defined \(u = x+2\). So we have: $$ \sinh^{-1}\left(\frac{3(x+2)-1}{1}\right) + C = \sinh^{-1}(3x + 5) + C $$ The final answer is: $$ \int \frac{(3x+5) dx}{\sqrt{x^2 + 4x + 3}} = \sinh^{-1}(3x + 5) + C $$

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