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Chapter 7: Problem 79

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}} ; x=u^{6}$$

Short Answer

Expert verified
Answer: The final expression for the integral is: $$\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}}=6[\frac{1}{3}(x^\frac{1}{6})^3-\frac{1}{2}(x^\frac{1}{6})^2+x^\frac{1}{6} -\ln{|x^\frac{1}{6}+1|}] +C$$

Step by step solution

01

Apply the substitution \(x=u^6\)

We have \(x=u^6\). Take the derivative of both sides with respect to \(u\) to find \(dx\). This gives: $$\frac{d x}{d u}=6 u^5$$ $$d x=6 u^5 d u$$ Now we'll substitute \(x=u^6\) and \(dx=6u^5du\) into the integral: $$\int \frac{d x}{\sqrt{x}+\sqrt[3]{x} } = \int \frac{6 u^5 d u}{\sqrt{u^6}+\sqrt[3]{u^6}}$$ #Step 2: Simplify the integral#
02

Simplify the integral

Now let's simplify the integral: $$\int \frac{6 u^5 d u}{\sqrt{u^6}+\sqrt[3]{u^6}}=\int \frac{6 u^5 d u}{u^3+u^2}$$ #Step 3: Make a substitution#
03

Divide both the numerator and denominator by \(u^2\)

Divide both the numerator and denominator by \(u^2\) to simplify the integral: $$\int \frac{6 u^5 d u}{u^3+u^2}=\int \frac{6 u^3 d u}{u+1}$$ #Step 4: Integrate using long division#
04

Perform long division to simplify the fraction

Divide the numerator by the denominator, which gives: $$u^3= (u+1)(u^2-u+1)+(-1)$$ So we have: $$\int \frac{6 u^3 d u}{u+1}=6\int \frac{(u+1)(u^2-u+1)-1}{u+1} du$$ #Step 5: Evaluate the integral#
05

Evaluate the integral

Now we can evaluate the integral as follows: $$6 \int \frac{(u+1)(u^2-u+1)-1}{u+1} d u=6 \int (u^2-u+1) d u -6 \int \frac{1}{u+1} d u$$ Now we can easily find the antiderivatives: $$6 \int (u^2-u+1) d u = 6 [\frac{1}{3}u^3-\frac{1}{2}u^2+u ] + C_1$$ $$-6 \int \frac{1}{u+1} d u=-6 \ln{|u+1|}+C_2$$ Add both antiderivatives: $$\int \frac{6 u^3 d u}{u+1}=6 [\frac{1}{3}u^3-\frac{1}{2}u^2+u -\ln{|u+1|}] +C$$ where C is the constant of integration. #Step 6: Back-substitute \(x=u^6\)#
06

Substitute \(x=u^6\) back into the expression

Finally, substitute \(x=u^6\) back into the expression to get the final answer: $$\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}}=6[\frac{1}{3}(x^\frac{1}{6})^3-\frac{1}{2}(x^\frac{1}{6})^2+x^\frac{1}{6} -\ln{|x^\frac{1}{6}+1|}] +C$$

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Most popular questions from this chapter

By reduction formula 4 in Section 3 $$\int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C$$ Graph the following functions and find the area under the curve on the given interval. $$f(x)=\left(x^{2}-25\right)^{1 / 2},[5,10]$$

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{3 x^{2}+4 x-6}{x^{2}-3 x+2} d x$$

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