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Chapter 4: Problem 41

Use a computer algebra system to find or evaluate the integral. $$ \int \frac{1}{1+\sqrt{x}} d x $$

Short Answer

Expert verified
The integral of \(\frac{1}{1+\sqrt{x}} dx\) is \(2 \ln | \frac{1}{\sqrt{x}} + 1| + C\).

Step by step solution

01

Perform the substitution

Let \(u = \sqrt{x}\). This means that \(x = u^2\). In terms of \(u\), \(dx\) becomes \(2udu\). Substituting these into the integral gives us: \(\int \frac{1}{1+u} * 2u du\).
02

Simplify the integral

The integral can be simplified by multiplying through the \(2u\): \(\int \frac{2u}{1+u} du\). This can be further simplified into a form that's easier to integrate by dividing each term in the numerator by \(u\). This gives: \(\int \frac{2}{\frac{1}{u}+1} du\).
03

Evaluate the integral

Now, we can straightforwardly integrate this function: \(\int \frac{2}{\frac{1}{u}+1} du = 2 \int \frac{1}{\frac{1}{u}+1} du = 2 \ln |\frac{1}{u} + 1| + C\). We've accounted for the constant of integration \(C\).
04

Convert the answer back to terms of \(x\)

Finally, we need to convert the answer back to terms of \(x\) by substituting \(u = \sqrt{x}\) back into the answer: \(2 \ln | \frac{1}{\sqrt{x}} + 1| + C\).

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Most popular questions from this chapter

Consider the integral \(\int \frac{1}{\sqrt{6 x-x^{2}}} d x\). (a) Find the integral by completing the square of the radicand. (b) Find the integral by making the substitution \(u=\sqrt{x}\). (c) The antiderivatives in parts (a) and (b) appear to be significantly different. Use a graphing utility to graph each antiderivative in the same viewing window and determine the relationship between them. Find the domain of each.

Solve the differential equation. \(\frac{d y}{d x}=\frac{1}{(x-1) \sqrt{-4 x^{2}+8 x-1}}\)

Solve the differential equation. \(\frac{d y}{d x}=\frac{x^{3}-21 x}{5+4 x-x^{2}}\)

Find the indefinite integral using the formulas of Theorem 4.24 \(\int \frac{d x}{(x+2) \sqrt{x^{2}+4 x+8}}\)

Show that if \(f\) is continuous on the entire real number line, then \(\int_{a}^{b} f(x+h) d x=\int_{a+h}^{b+h} f(x) d x\)
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