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Chapter 4: Problem 15

Find the indefinite integral by \(u\) -substitution. (Hint: Let \(u\) be the denominator of the integrand.) $$ \int \frac{1}{1+\sqrt{2 x}} d x $$

Short Answer

Expert verified
The antiderivative, or the indefinite integral of \( \frac{1}{1+\sqrt{2 x}} \) is \( (1 + \sqrt{2x})^2 -2( 1+ \sqrt{2x}) +c\).

Step by step solution

01

Define 'u'

Set \(u\) equal to the denominator of the integrand. Thus, \(u = 1 + \sqrt{2x}\).
02

Differentiate 'u'

Now take the derivative of \(u\) with respect to \(x\). The derivative, \(du/dx\), is \(du/dx = d(1 + \sqrt{2x})/dx = \sqrt{2}/2\sqrt{x}\). Multiply both sides by \(dx\) to isolate \(du\). So, \(du = \sqrt{2}/2\sqrt{x} dx\). Multiply both sides of the equation by \(2\sqrt{x}\) to isolate \(dx\). So, \(dx = 2\sqrt{x} du\).
03

Substitute 'u' and 'dx' into the equation

Now substitute \(u\) and \(dx\) into the integral. The integral thus becomes: \(\int \frac{1}{u} \cdot 2\sqrt{x} du\).
04

Substitute 'x' using original 'u' statement

Now we have to eliminate the \(x\) in the integrand. From the equation of \(u = 1 + \sqrt{2x}\), we can solve for \(x\). Which is \(x = (u-1)^2/2\). Substituting \(x\) in the integrand, we get: \(\int \frac{1}{u} \cdot 2\sqrt{(u-1)^2/2} du\).
05

Simplify and Integrate

The integrand simplifies to \( 2(u-1)\), leading to the integral of \( 2(u-1) du\). The integral of \(2(u-1)\) with respect to \(u\) is \(u^2-2u+c\).
06

Replace 'u'

Now, replace \(u\) with \(1 + \sqrt{2x}\), the original substitution. This yields: \( (1 + \sqrt{2x})^2 -2( 1+ \sqrt{2x}) +c\). This is the final solution.

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