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This question does not require integration by parts.

\int_{0}^{\frac{\pi}{4}}\left(\sin^{-1}\left(\tan x\right)+\tan^{-1}\left(\sin2x\right)+\tan^{-1}\left(\cos2x\right)\right)dx

\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(\sec^{2}x\right)\left(\sin^{-1}\left(\tan\left(\sin^{-1}\left(\frac{\tan\left(\sin^{-1}\sqrt{\frac{3}{4\tan x}}\right)}{\sqrt{6}}\right)\right)\right)\right)^{3}dx Feel free to share your attempt.:cool:

It's not my approach but you may try if it also works.;)

\int_{\frac{1}{e^{3}}}^{e^{3}}\ln\left(x\right)\cdot\frac{\ln\left(x+\sqrt{x}\right)+\sqrt{x}\ln\left(1+\sqrt{x}\right)}{x\left(1+\sqrt{x}\right)}dx

Yes, there is a removable singularity at x=0. If you want to do it more rigourously, you may treat the integrand as a piecewise continuous function.

\int_{-1}^{1}\left(\frac{\sin^{-1}x}{x}\right)^{3}\left(\frac{\pi^{x}}{1+\pi^{x}}\right)dx

\int_{0}^{\pi}\frac{x\sin^{4}x+\left(\sin2x\right)\log_{2}\left(2^{\pi}+4^{x}\right)}{3+\cos4x}dx

\int_{-10}^{10}\frac{2x^{2}\left(\sin^{4}x+\cos^{4}x\right)-x\sin4x+2}{\left(1+10^{\sin x}\right)\left(x^{2}\cos^{2}x-x\sin2x+1\right)\left(x^{2}\sin^{2}x+x\sin2x+1\right)}dx=11\pi+2\tan^{-1}\left(\frac{10\sin20+\cos20}{49\sin20+10\cos20}\right)

\int_{0}^{2}\left(\sqrt{4^{x+1}-8^{x}}\sqrt{4^{x}-5\left(2^{x}\right)+7}-\sqrt[3]{16^{x}}\sqrt[3]{2-2^{x}}\right)dx=\frac{6}{\ln2}

By the way, choosing the correct substitution is not the end of story.😈

You choose the correct substitution but calculate du/dx incorrectly. \frac{du}{dx}=\frac{2x^3-2x\cos^2(2\ln x)}{x^4+x^2\sin(4\ln x)+\cos^2(2\ln x)}

\int_{1}^{100}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx=\frac{11\pi}{8}+\frac{1}{2}\tan^{-1}\left(\frac{1+10000\tan\left(4\ln10\right)}{10000}\right)

In this case, it might be easier to think of it as "complementary". Changing theta in (c) to pi/2-theta gives the answer quickly.

You can do it in reverse (from bottom to top). f(pi/e -1)>f(0) e^(pi/e -1)-(pi/e -1)+1>0 e^(pi/e -1)>pi/e e^(pi/e)>pi pi/e>ln pi pi ln e>e ln pi e^pi>pi^e

By adjusting the coefficient of k to 1, you should be able to take out log_2 (2), which is just 1. Integrating dx from 0 to 1 gives you 1. The remaining factors have the same structure and you can offset them by suitable substitution.

What tricks have you tried? The following techniques are NOT required. 😈 integration by parts differentiation under the integral sign hyperbolic function

The integral is 1 irrespective of the value of k.

I wrote this in the past but I have now forgotten how to solve it...

I made a typo in Wolfram earlier. It does give a nice answer. \int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}