Question

give me some super hard derivatives, integrations wit answers

Answer 1

I'll give you some tricky ones but not hard ones
\[\int\limits \frac{1}{1+\sqrt{x}}dx \\ \\ \int\limits \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx \\ \\ \int\limits \cos(\ln(x))dx \\ \\ \int\limits \frac{1}{1+\sin(2x)}dx\]

Answer 2

Answer
\[2\sqrt{x}-2\ln(1+\sqrt{x}) \\ \\ 2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\ln(1+x^{\frac{1}{6}})\]
\[\frac{[\cos(lnx)+\sin(lnx)]x}{2} \\ \\ -\frac{1}{1+tanx}\]

Answer 3

The integration part will be different

Answer 4

Here's something interesting
\[\LARGE \int2\sec^2x \ \tan \ x dx\]

Answer 5

Depending on how you substitute, you may get one of two integrals...
\[\LARGE \sec^2x\]
OR
\[\LARGE \tan^2x\]

Answer 6

Two different integrals? What is going on here? ^.^

Answer 7

This is a striking example of why it is important not to leave the +C out.
Because
\[\LARGE \sec^2x=\tan^2x+1\]
So,
\[\LARGE \sec^2x \ + \ C=\tan^2 x \ + \ 1 + \ C\]
but since 1 itself is a constant, the C sort of "swallows" up the 1. So...
\[\LARGE \int2\sec^2x \ \tan \ x dx=\sec^2x \ + \ C = \tan^2x \ + \ C\]

Answer 8

wow good ones....

Answer 9

For the one @terenzreignz gave, you can use substitution for to make things easier as well as use the trig identity.\[\sec^2x=\tan^2x+1\]\[\int\limits_{}^{}2\sec^2(x)\tan(x)dx =\int\limits_{}^{}2(\tan^2(x)+1)\tan(x)dx \rightarrow u = \tan(x) \rightarrow \frac{ du }{ \sec^2(x) }=dx\]\[=\int\limits_{}^{}[2(u^2+1)u]\frac{ du }{ \sec^2(x) }\]At this point, we can rewrite sec^2(x) as u^2 + 1 since we have defined u as tan(x).\[=\int\limits_{}^{}[2\cancel{(u^2+1)}u]\frac{ du }{ \cancel{u^2+1 }} =\int\limits_{}^{}(2u)du=u^2+C=\tan^2(x)+C\]Or do it straight away the way he did and get the answer as sec^2(x) + C. There isn't much of a mystery other than like he said that it gets eaten up by the C.

Answer 10

\[\begin{align*}
\int\limits_0^1u^x(\ln u)^n\,\mathrm du\qquad&n\in\mathbb Z>0\\\hline
\text{let } u =e^{-w}\\
\mathrm du=-e^{-w}\mathrm dw\\
u=0\rightarrow w=\infty\\
u=1\rightarrow w=0\\
&=\int\limits_\infty^0e^{-xw}(-w)^n(-e^{-w})\mathrm dw\\
&=(-1)^{n}\int\limits_0^\infty w^ne^{-(x+1)w}\mathrm dw\\
&=(-1)^{n}\left[\left.\frac{w^ne^{-(x+1)w}}{-(x+1)}\right|_0^\infty-\int\limits_0^\infty \frac{nw^{n-1}e^{-(x+1)w}}{-(x+1)}\mathrm dw\right]\\
&=(-1)^{n}\left[0+\frac{n}{x+1}\int\limits_0^\infty w^{n-1}e^{-(x+1)w}\mathrm dw\right]\\
&=\frac{n(-1)^{n}}{x+1}\int\limits_0^\infty w^{n-1}e^{-(x+1)w}\mathrm dw\\
&=\frac{n(-1)^{n}}{x+1}\left[\left.\frac{w^{n-1}e^{-(x+1)w}}{-(x+1)}\right|_0^\infty-\int\limits_0^\infty (n-1)w^{n-2}\frac{e^{-(x+1)w}}{-(x+1)}\mathrm dw\right]\\
&=\frac{n(-1)^{n}}{x+1}\left[0+\frac{n-1}{x+1}\int\limits_0^\infty w^{n-2}e^{-(x+1)w}\mathrm dw\right]\\
&=\frac{n(n-1)(-1)^{n}}{(x+1)^2}\int\limits_0^\infty w^{n-2}e^{-(x+1)w}\mathrm dw\\
&=\frac{n(n-1)(-1)^{n}}{(x+1)^2}\left[\left.\frac{w^{n-2}w^{-(x+1)w}}{-(x+1)}\right|_0^\infty-\int\limits_0^\infty\frac{(n-2)w^{n-3}w^{-(x+1)w}}{-(x+1)}\mathrm dw\right]\\
&=\frac{n(n-1)(-1)^{n}}{(x+1)^2}\left[0+\frac{n-2}{x+1}\int\limits_0^\infty w^{n-3}w^{-(x+1)w}\mathrm dw\right]\\
&=\frac{n(n-1)(n-2)(-1)^{n}}{(x+1)^3}\int\limits_0^\infty w^{n-3}w^{-(x+1)w}\mathrm dw\\
&=\qquad\vdots\\
\\
&=\frac{n!(-1)^{n}}{(x+1)^n}
\end{align*}
\]