Copyright

(1)Evaluate the indefinite integral. (Use C for the constant of integration.) { \int\frac{ 9 +...

Question:

(1)Evaluate the indefinite integral. (Use C for the constant of integration.)

{eq}\int\frac{ 9 + 7x}{1 + x^2 }dx {/eq}

_____

(2)Evaluate the indefinite integral. (Use C for the constant of integration.)

{eq}\int x(5 - x)^9 dx {/eq}

_____

Evaluate the indefinite integral. (Use C for the constant of integration.)

{eq}\int x^3 \sqrt{ x^2 + 36} dx {/eq}

_____

Indefinite Integrals.

Indefinite integrals are evaluated as fundamental theorem of calculus.

Here we substitute the value by u-substitution method to just change the variable of one form to make it easier.

The formula is:

{eq}\displaystyle\int x^n\ dx=\dfrac{x^{n+1}}{n+1}+c\\\\ \displaystyle\int \dfrac{1}{x^2+1}dx=tan^{-1}(x)+c\\\\ {/eq}

Answer and Explanation:

We have to calculate the given indefinite integrals using c as an integration constant.

Part 1.)

{eq}\displaystyle\int \dfrac{9+7x}{1+x^2}dx\\\\ {/eq}

Separating it:

{eq}=\displaystyle\int \dfrac{9}{1+x^2}dx+\displaystyle\int \dfrac{7x}{1+x^2}dx\\\\ {/eq}

Substituting:

{eq}1+x^2=z\\\\ 2x\ dx=dz\\\\ x\ dx=\dfrac{dz}{2}\\\\ {/eq}

Therefore:

{eq}=9tan^{-1}(x)+7\displaystyle\int \dfrac{1}{z}\cdot \dfrac{dz}{2}\\\\ =9tan^{-1}(x)+\dfrac{7}{2}\left [ ln\left | z \right | \right ]+c\\\\ =9tan^{-1}(x)+\dfrac{7}{2}\left [ ln\left | 1+x^2 \right | \right ]+c\\\\ {/eq}

Here c defines the integration constant.

Part 2.)

{eq}\displaystyle\int x(5-x)^9\ dx\\\\ {/eq}

Substituting:

{eq}5-x=z\\\\ -dx=dz\\\\ dx=-dz\\\\ {/eq}

And {eq}x=5-z\\\\ {/eq}

Therefore:

{eq}=\displaystyle\int (5-z)z^9(-dz)\\\\ =-\displaystyle\int 5z^9\ dz+\displaystyle\int z^{10}dz\\\\ =-5\left [ \dfrac{z^{10}}{10} \right ]+\left [ \dfrac{z^{11}}{11} \right ]+c\\\\ =\dfrac{-1}{2}(5-x)^{10}+\dfrac{1}{11}(5-x)^{11}+c\\\\ {/eq}

Here c defines integration constant.

Part 3.)

{eq}\displaystyle\int x^3\sqrt {x^2+36}dx\\\\ =\displaystyle\int x^2\cdot x\sqrt {x^2+36}dx\\\\ {/eq}

Substituting:

{eq}x^2+36=z\\\\ 2x\ dx=dz\\\\ x\ dx=\dfrac{dz}{2}\\\\ {/eq}

Or,

{eq}x^2=z-36\\\\ {/eq}

Therefore:

{eq}=\displaystyle\int (z-36)\sqrt z\cdot \dfrac{dz}{2}\\\\ =\displaystyle\int z\cdot \sqrt z\cdot \dfrac{dz}{2}-\displaystyle\int 36\sqrt z\cdot \dfrac{dz}{2}\\\\ =\dfrac{1}{2}\displaystyle\int z^{\frac{3}{2}}dz-18\displaystyle\int z^{\frac{1}{2}}dz\\\\ =\dfrac{1}{2}\left [ \dfrac{z^{\frac{5}{2}}}{\frac{5}{2}} \right ]-18\left [ \dfrac{z^{\frac{3}{2}}}{\frac{3}{2}} \right ]+c\\\\ =\dfrac{1}{2}\cdot \dfrac{2}{5}\left [ x^2+36 \right ]^{\frac{5}{2}}-18\cdot \dfrac{2}{3}\left [ x^2+36 \right ]^{\frac{3}{2}}+c\\\\ =\dfrac{1}{5}\left [ x^2+36 \right ]^{\frac{5}{2}}-12\left [ x^2+36 \right ]^{\frac{3}{2}}+c\\\\ {/eq}

Here c defines the integration constant.


Learn more about this topic:

Loading...
Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
113K

Related to this Question

Explore our homework questions and answers library