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All Calculus AB Resources

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Flashcards: Use Substitution and Integration

Use a change of variable (aka a u-substitution) to evaluate the integral,

$\int x\sqrt{1+x^2}dx$

$\frac{3}{4}(x^2+1)^{\frac{3}{2}}+C$

$\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C$

$\frac{1}{6}(x^2+1)^{\frac{2}{3}}+\sqrt{x}+C$

$\frac{1}{6}(2x+1)^{\frac{2}{3}}+C$

$\frac{1}{3}x^{\frac{3}{2}}+C$

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Flashcards: Use Substitution and Integration

Use a change of variable (aka a u-substitution) to evaluate the integral,

$\int x\sqrt{1+x^2}dx$

Your answer:

$\frac{3}{4}(x^2+1)^{\frac{3}{2}}+C$

Your answer:

$\frac{1}{6}(x^2+1)^{\frac{2}{3}}+\sqrt{x}+C$

Your answer:

$\frac{1}{6}(2x+1)^{\frac{2}{3}}+C$

Your answer:

$\frac{1}{3}x^{\frac{3}{2}}+C$

Correct answer:

$\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C$

Explanation:

$\int x\sqrt{1+x^2}dx$

Integrals such as this are seen very commonly in introductory calculus courses. It is often useful to look for patterns such as the fact that the polynomial under the radical in our example, 1+x^2 , happens to be one order higher than the factor outside the radical, You know that if you take a derivative of a second order polynomial you will get a first order polynomial, so let's define the variable:

u = 1+x^2 (1)

Now differentiate with respect to to write the differential for ,

du = 2x dx (2)

Looking at equation (2), we can solve for xdx , to obtain $xdx =\frac{1}{2}du$ . Now if we look at the original integral we can rewrite in terms of

$\int x\sqrt{1+x^2}dx = \int\sqrt{1+x^2}\underbrace{xdx} = \int\sqrt{u}\left(\frac{1}{2}du \right )$

Now proceed with the integration with respect to .

$\int\sqrt{u}\left(\frac{1}{2}du \right ) = \frac{1}{2}\int u^{\frac{1}{2}}du$

$= \frac{1}{2}\left[\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1} \right ]+C$

$=\frac{1}{2} \left(\frac{2}{3}u^{\frac{3}{2}} \right )+C$

$=\frac{1}{3}u^{\frac{3}{2}}+C$

Now write the result in terms of using equation (1), we conclude,

$\int x\sqrt{1+x^2}dx=\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C$

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All Calculus AB Resources

Flashcards: Use Substitution and Integration

Flashcards: Use Substitution and Integration

Flashcards: Use Substitution and Integration

Flashcards: Use Substitution and Integration

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All Calculus AB Resources

Flashcards: Use Substitution and Integration

Flashcards: Use Substitution and Integration

Flashcards: Use Substitution and Integration

Flashcards: Use Substitution and Integration

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