Use a table of integrals to find the indefinite integral. $$ \int x^{2} \sqrt{12+25 x^{2}} d x $$

Question

UpStudy AI · Accepted Answer

We wish to evaluate

$$
\int x^2 \sqrt{12+25 x^2}\,dx.
$$

A good idea is to first change the variable so that the constant 25 is absorbed. Let

$$
u=5x.
$$

Then

$$
x=\frac{u}{5},\quad x^2=\frac{u^2}{25},\quad dx=\frac{du}{5},
$$

and

$$
\sqrt{12+25x^2}=\sqrt{12+u^2}\,.
$$

Thus the integral becomes

$$
\int x^2 \sqrt{12+25x^2}\,dx=\int \frac{u^2}{25}\sqrt{12+u^2}\cdot\frac{du}{5}=\frac{1}{125}\int u^2\sqrt{12+u^2}\,du\,.
$$

Let us denote

$$
I=\int u^2 \sqrt{12+u^2}\,du\,.
$$

One may show (for example, by consulting an integral table) that for a general constant $$a^2$$ the following formula holds:

$$
\int x^2\sqrt{x^2+a^2}\,dx=\frac{x\,\sqrt{x^2+a^2}\,\bigl(2x^2-a^2\bigr)}{8}+\frac{a^4}{8}\ln\Big|x+\sqrt{x^2+a^2}\,\Big|+C\,.
$$

In our case we have $$a^2=12$$ (so that $$a=\sqrt{12}=2\sqrt{3}$$) and the variable is $$u$$. Thus we get

$$
I=\frac{u\,\sqrt{u^2+12}\,\Bigl(2u^2-12\Bigr)}{8}+\frac{12^2}{8}\ln\Big|u+\sqrt{u^2+12}\,\Big|+C\,.
$$

Since $$12^2=144$$, this becomes

$$
I=\frac{u\,\sqrt{u^2+12}\,(2u^2-12)}{8}+18\,\ln\Big|u+\sqrt{u^2+12}\,\Big|+C\,.
$$

Recalling that the original integral in terms of $$x$$ was

$$
\int x^2\sqrt{12+25x^2}\,dx=\frac{1}{125}\,I\,,
$$

and substituting back $$u=5x$$, we obtain

$$
\int x^2\sqrt{12+25x^2}\,dx=\frac{1}{125}\left[\frac{5x\,\sqrt{12+25x^2}\Bigl(2(5x)^2-12\Bigr)}{8}+18\,\ln\Big|5x+\sqrt{12+25x^2}\,\Big|\right]+C\,.
$$

Notice that

$$
2(5x)^2-12=2(25x^2)-12=50x^2-12\,,
$$

so that

$$
\frac{5x(50x^2-12)}{8}=\frac{250x^3-60x}{8}\,.
$$

Thus we can write

$$
\int x^2\sqrt{12+25x^2}\,dx=\frac{1}{125}\left[\frac{250x^3-60x}{8}\sqrt{12+25x^2}+18\,\ln\Big|5x+\sqrt{12+25x^2}\,\Big|\right]+C\,.
$$

We may simplify the constants in the first term:

$$
\frac{1}{125}\cdot\frac{250}{8}=\frac{250}{1000}=\frac{1}{4}\,,
$$

and similarly for the second term the constant becomes $$\frac{18}{125}$$. Therefore, our final answer is

$$
\boxed{\frac{x(25x^2-6)\sqrt{12+25x^2}}{4}+\frac{18}{125}\ln\Big|5x+\sqrt{12+25x^2}\,\Big|+C}\,.
$$
$$ \int x^{2} \sqrt{12+25 x^{2}} \, dx = \frac{x(25x^{2} - 6)\sqrt{12+25x^{2}}}{4} + \frac{18}{125} \ln\left|5x + \sqrt{12+25x^{2}}\right| + C $$

Use a table of integrals to find the indefinite integral.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Calculus Questions