I have some integrals to solve for later...

[pantentic]

I have some integrals to solve for later. But the thing is, my math knowledge is meager, so if you could help me out I would be grateful

\(\displaystyle a)\, \lim_{x\, \rightarrow\, \infty}\, \left(\, \dfrac{2x\, +\, k}{2x\, -\, b}\, \right)^{kx + b}\)

\(\displaystyle b)\, \lim_{x\, \rightarrow\, \infty}\, \left(\, \sqrt[3]{\strut x^3\, +\, 7x\,}\, -\, x\, \right)\)

\(\displaystyle c)\, \int\, \dfrac{x\, (1\, -\, x)}{x^2\, +\, 1}\, dx\)

\(\displaystyle d)\, \int\, \left[\, x^2\, \arctan(x^3\, +\, 7)\, \right]\, dx\)

\(\displaystyle e)\, \int_0^{\infty}\, \cos(2x)\, dx\)

\(\displaystyle f)\, \int_{-\infty}^{+\infty}\, \dfrac{1}{a^2\, +\, x^2}\, dx\)

 

[Deleted member 4993]

But the thing is, my math knowledge is meager, so if you could help me out I would be grateful

\(\displaystyle a)\, \lim_{x\, \rightarrow\, \infty}\, \left(\, \dfrac{2x\, +\, k}{2x\, -\, b}\, \right)^{kx + b}\)

\(\displaystyle b)\, \lim_{x\, \rightarrow\, \infty}\, \left(\, \sqrt[3]{\strut x^3\, +\, 7x\,}\, -\, x\, \right)\)

\(\displaystyle c)\, \int\, \dfrac{x\, (1\, -\, x)}{x^2\, +\, 1}\, dx\)

\(\displaystyle d)\, \int\, \left[\, x^2\, \arctan(x^3\, +\, 7)\, \right]\, dx\)

\(\displaystyle e)\, \int_0^{\infty}\, \cos(2x)\, dx\)

\(\displaystyle f)\, \int_{-\infty}^{+\infty}\, \dfrac{1}{a^2\, +\, x^2}\, dx\)

What are your thoughts?

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[Prove It]

\(\displaystyle a)\, \lim_{x\, \rightarrow\, \infty}\, \left(\, \dfrac{2x\, +\, k}{2x\, -\, b}\, \right)^{kx + b}\)

\(\displaystyle b)\, \lim_{x\, \rightarrow\, \infty}\, \left(\, \sqrt[3]{\strut x^3\, +\, 7x\,}\, -\, x\, \right)\)

\(\displaystyle c)\, \int\, \dfrac{x\, (1\, -\, x)}{x^2\, +\, 1}\, dx\)

\(\displaystyle d)\, \int\, \left[\, x^2\, \arctan(x^3\, +\, 7)\, \right]\, dx\)

\(\displaystyle e)\, \int_0^{\infty}\, \cos(2x)\, dx\)

\(\displaystyle f)\, \int_{-\infty}^{+\infty}\, \dfrac{1}{a^2\, +\, x^2}\, dx\)

a) Writing it as \(\displaystyle \displaystyle \begin{align*} \mathrm{e}^{ \ln{ \left[ \left( \frac{2\,x + k}{2\,x - b} \right) ^{ k\,x + b } \right] } } \end{align*}\) and then applying some logarithm laws will help simplify it to a form where you can use L'Hospital's Rule.

b) I would try rationalising the numerator. The difference of two cubes rule will help.

c) Long divide.

d) \(\displaystyle \displaystyle \begin{align*} \int{ x^2\arctan{ \left( x^3 + 7 \right) } \,\mathrm{d}x } &= \frac{1}{3} \int{ 3\,x^2\arctan{ \left( x^3 + 7 \right) } \,\mathrm{d}x } \\ &= \frac{1}{3} \int{ \arctan{ \left( u \right) } \,\mathrm{d}u } \textrm{ after substituting } u = x^3 + 7 \implies \mathrm{d}u = 3\,x^2\,\mathrm{d}x \end{align*}\)

Do you know how to find the integral of the arctangent function?

e) Is it possible to evaluate the limit of a sine function at infinity?

f) What sort of an integral will this form be?

 

[Berationalgetreal]

a) Writing it as \(\displaystyle \displaystyle \begin{align*} \mathrm{e}^{ \ln{ \left[ \left( \frac{2\,x + k}{2\,x - b} \right) ^{ k\,x + b } \right] } } \end{align*}\) and then applying some logarithm laws will help simplify it to a form where you can use L'Hospital's Rule.

b) I would try rationalising the numerator. The difference of two cubes rule will help.

c) Long divide.

d) \(\displaystyle \displaystyle \begin{align*} \int{ x^2\arctan{ \left( x^3 + 7 \right) } \,\mathrm{d}x } &= \frac{1}{3} \int{ 3\,x^2\arctan{ \left( x^3 + 7 \right) } \,\mathrm{d}x } \\ &= \frac{1}{3} \int{ \arctan{ \left( u \right) } \,\mathrm{d}u } \textrm{ after substituting } u = x^3 + 7 \implies \mathrm{d}u = 3\,x^2\,\mathrm{d}x \end{align*}\)

Do you know how to find the integral of the arctangent function?

e) Is it possible to evaluate the limit of a sine function at infinity?

f) What sort of an integral will this form be?

For a) I suggest not to waste time with De L'Hopital's rule. This exercise can be easily solved writing

\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, +\infty}\, \)\(\displaystyle \bigg(\, 1\, +\, \left(\, \dfrac{2x\, +\, k}{2x\, -\, b}\, -\, 1\, \right)\, \bigg)^{kx\, +\, b}\)

And then using e's limit:

\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, +\infty}\, \)\(\displaystyle \bigg(\, \left(\, 1\, +\, \dfrac{b\, +\, k}{2x\, -\, b}\, \right)^{\frac{2x\, -\, b}{b\, +\, k}}\, \bigg)^{\left( \frac{kx + b}{2x - b} \right)\cdot(\kappa + b)}\, = \, e^{\frac{k}{2}(k + b)}\)

 

[Ishuda]

It was here before so I didn't post anything. Seems gone now, so
\(\displaystyle \left(\, \frac{2x+k}{2x-b}\right)^{kx+b}\)
=\(\displaystyle \left(\, 1\, +\, ( \frac{2x+k}{2x-b}\, -\, 1)\right)^{kx+b}\)
=\(\displaystyle \left(\, 1\, +\, ( \frac{k+b}{2x-b})\right)^{kx+b}\)
=\(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{kx+b}\)
=\(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{kx}\) \(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{b}\)
=\(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{\frac{2x-b}{k+b}\, \frac{k(k+b}{2}}\) \(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{\frac{b(2+k)}{2}}\)

Now let
t = \(\displaystyle \frac{2x-b}{k+b}\)
to get
\(\displaystyle \left(\, \frac{2x+k}{2x-b}\right)^{kx+b}\)
=\(\displaystyle \left(1\, +\, \frac{1}{t}\right)^{t\, \frac{k(k+b)}{2}}\) \(\displaystyle \left(\, 1\, +\, \frac{1}{t}\right)^{\frac{b(2+k)}{2}}\)
consider the various cases for k+b less than zero, greater than zero and equal to zero and use
\(\displaystyle \underset{x\, \to\, \infty}{lim}\, (1\, +\, \frac{1}{x})^x\, =\, e\)

 

[Berationalgetreal]

It was here before so I didn't post anything. Seems gone now, so
\(\displaystyle \left(\, \frac{2x+k}{2x-b}\right)^{kx+b}\)
=\(\displaystyle \left(\, 1\, +\, ( \frac{2x+k}{2x-b}\, -\, 1)\right)^{kx+b}\)
=\(\displaystyle \left(\, 1\, +\, ( \frac{k+b}{2x-b})\right)^{kx+b}\)
=\(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{kx+b}\)
=\(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{kx}\) \(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{b}\)
=\(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{\frac{2x-b}{k+b}\, \frac{k(k+b}{2}}\) \(\displaystyle \left(\, 1\, +\, \frac{1}{\frac{2x-b}{k+b}}\right)^{\frac{b(2+k)}{2}}\)

Now let
t = \(\displaystyle \frac{2x-b}{k+b}\)
to get
\(\displaystyle \left(\, \frac{2x+k}{2x-b}\right)^{kx+b}\)
=\(\displaystyle \left(1\, +\, \frac{1}{t}\right)^{t\, \frac{k(k+b)}{2}}\) \(\displaystyle \left(\, 1\, +\, \frac{1}{t}\right)^{\frac{b(2+k)}{2}}\)
consider the various cases for k+b less than zero, greater than zero and equal to zero and use
\(\displaystyle \underset{x\, \to\, \infty}{lim}\, (1\, +\, \frac{1}{x})^x\, =\, e\)

Yeah, I don't know why but my answer kind of got deleted.