/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 11

Evaluate the following integrals. $$\int \frac{x}{\sqrt{x+1}} d x$$

Short Answer

Expert verified
Question: Evaluate the integral \(\int \frac{x}{\sqrt{x+1}} d x\). Answer: \(\frac{2}{3}(x+1)^{\frac{3}{2}} - 2(x+1)^{\frac{1}{2}} + C\)

Step by step solution

01

Now, let's compute the differential, \(du\). #Step 2 - Find differential# We will differentiate the substituted variable with respect to x. Differentiating \(u = x + 1\) with respect to \(x\), we get

\( \frac{d u}{d x} = \frac{d (x+1)}{d x} = 1\). So \(du = dx\). #Step 3 - Substitute variable# To rewrite the integral in terms of \(u\), we need to substitute not only the variable \(x\) and \(dx\) but also the limits of integration. Now, we have:
02

\(u = x+1\) and \(du = dx\). With these substitutions, we get: $$\int \frac{x}{\sqrt{x+1}} d x = \int \frac{u - 1}{\sqrt{u}} d u$$ #Step 4 - Simplify the integral# Next, we divide each term by \(u^{\frac{1}{2}}\):

$$\int \frac{u - 1}{\sqrt{u}} d u = \int (u^{\frac{1}{2}} - u^{-\frac{1}{2}}) d u$$ #Step 5 - Perform integration# Now, we can integrate term by term:
03

$$\int (u^{\frac{1}{2}} - u^{-\frac{1}{2}}) d u = \frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}} + C$$ #Step 6 - Substitute back for x# Finally, substitute back the original variable \(x\) by using \(u = x+1\):

$$\frac{2}{3}(x+1)^{\frac{3}{2}} - 2(x+1)^{\frac{1}{2}} + C$$ And so, the evaluated integral is: $$\int \frac{x}{\sqrt{x+1}} d x = \frac{2}{3}(x+1)^{\frac{3}{2}} - 2(x+1)^{\frac{1}{2}} + C$$

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Most popular questions from this chapter

Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b],\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 2 and an illustration.

Graph the function \(f(x)=\frac{1}{x \sqrt{x^{2}-36}}\) on its domain. Then find the area of the region \(R_{1}\) bounded by the curve and the \(x\) -axis on \([-12,-12 / \sqrt{3}]\) and the area of the region \(R_{2}\) bounded by the curve and the \(x\) -axis on \([12 / \sqrt{3}, 12] .\) Be sure your results are consistent with the graph.

Consider the curve \(y=\ln x\) a. Find the length of the curve from \(x=1\) to \(x=a\) and call it \(L(a) .\) (Hint: The change of variables \(u=\sqrt{x^{2}+1}\) allows evaluation by partial fractions.) b. Graph \(L(a)\) c. As \(a\) increases, \(L(a)\) increases as what power of \(a ?\)

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