I have this integral expression to be evaluated:     $$\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}$$   I've followed these steps:   Completed the square in $x$ and then substituted $u=x-\frac{1}{2}.$ Trig substituted $u = \frac{\sqrt{3}}{2}\tan t.$ and then simplified the equation thereby giving:    $$\int{\frac{\sqrt{3}\sec^2t}{\sqrt{3}\tan t+\sqrt{3}\sec t+1}dt}$$   I know up to this, but I'm unclear what to do next. Feel free to correct me if I'm wrong.
123971
Evaluate :  $$\frac{dx}{x+ \sqrt{x^2-x+1}}$$  After dividing and multiply with $x-\sqrt{x^2-x+1}$, I get  $x+\ln |x-1|-\int \frac{\sqrt{x^2-x+1}}{x-1}dx$.  Is $\int \frac{\sqrt{x^2-x+1}}{x-1}dx$ is integrable in terms of elementary functions?
102137
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$   My approach :   Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then  $$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$  But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
98396
Can you please provide any sort of hint or suggestion on how to find the following indefinite integral?  $$\int\frac{x^3}{\sqrt{x^2+1}}\text{d}x$$  I tried substituting everything but it didn't work. I also tried trigonometric substitution but I couldn't find any valid trigonometric identity for $ \sqrt{\cos^2(x)+1} $ or $ \sqrt{\sin^2(x)+1} $.
40780
Evaluate: $$\int_{-\pi/4}^{\pi/4}(\cos(t)+\sqrt{1+t^2}\cos(t)^3\sin(t)^3)dt$$  I have tried substituting $t=\tan(\theta)$ and breaking up the $\cos(t)^3$ using $\cos(t)^2=1-\sin(t)^2$, but I can't figure it out.  This question is #21 from the Math subject GRE form GR1268.
60553
Calculate the following integral by hands: $$\int_{0}^{\frac{\pi}{2}}\frac{\sin^{3}(t)}{\sin^{3}(t)+\cos^{3}(t)}dt$$  It seems apparently that integration by part does not work. I think integration by substitution could work but I can not figure out how to substitute.  I have calculated it by Wolfram and the result is $\frac{\pi}{4}$.
42803
The original question was to evaluate:   $$\int_{0}^{1} \frac{1-x}{(1+x) \ln x}\,dx$$  Using the substitution $x=\tan^2 \theta $, I simplified it down to the integral $\int_{0}^{\frac{\pi}{4}} \frac{\tan \theta - \tan^3 \theta}{\ln \tan \theta}$.  From here, I am stuck and am not sure where to continue.
94094
$$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$ How to integrate? I tried the substitution $x=\sin\theta$ but didn't work.
70577
Compute the indefinite integral   $$ \int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}} $$   My Attempt:  $$ \int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}=\int\frac{1}{(x^2-x+1)^{3/2}.\sqrt{\dfrac{x^2+x+1}{x^2-x+1}}}\,dx $$  Now define $t$ such that $t^2=\dfrac{x^2+x+1}{x^2-x+1}$ to get  $$ \begin{align} 2t\,dt &amp;= \frac{(x^2-x+1)(2x+1)-(x^2+x+1)\cdot (2x-1)}{(x^2-x+1)^2}\,dx\\ 2tdt &amp;= \frac{-4x^2+2x+2}{(x^2-x+1)^2}dx \end{align} $$  I don't know how to proceed from here.
115290
\begin{eqnarray*} \int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx &amp;=&amp;  \int \frac{(3\sin x+2\cos x)/\cos x}{(2\sin x+3\cos x)/\cos x}dx\\ \\ &amp;=&amp; \int \frac{3\tan x +2}{2\tan x +3} dx\\ &amp;&amp; u = \tan x \text{ and } du = \sec^2 x \ dx  \end{eqnarray*} Am I going in the right direction with this one? It seems like not.
117001
Evaluate $$\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}d x$$ What substitution should I use ? $\sqrt\frac{1+x}{1-x}$ suggests $x=cos2\theta $ but its not useful in $\frac{2-3x}{2+3x}$
19538
Evaluate $$\int \frac{1}{\sin x+\sec x}\,dx $$ Expressing $\sin x$ and $\cos x$ in terms of $\tan\frac{x}{2}$ i.e. putting $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$ and hence $dx=\dfrac{2\,dt}{1+t^2}$ $$\int \frac{1-t^2}{1+t-t^3}\,dt $$
69250
$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$  Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$  $\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different way to solve it?
65533
I was having a bit of trouble with an integral  $$\int {1 \over (x^2-1)^2}dx$$  It was given as a partial fractions problem, and I tried a substitution of  $x=sec(\theta);dx=sec(\theta)tan(\theta)d\theta$  to get $$\int{{\sec(\theta) \tan(\theta)}\over{\tan^4(\theta)}}d\theta=\int{\cos^2(\theta)\over \sin^3(\theta)}d\theta$$  but I can't seem to move past this. Any suggestions would be greatly appreciated.
78332
Evaluate   $$\int\sec^4(u) \operatorname d \!u$$  I don't know what to substitute: I've tried $1+\tan(u)$ and integration by parts. I know the general formula for $\sec^n(u)$, but I want to be able to do this integral on my own.
47025
$$\int x^3\sqrt{4-x^2}dx$$ I tried changing the expression like this: $$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$ And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
37037
How do I solve evaluate $$\int \frac{\mathrm{d}x}{e^x + 1}\ ?$$  I know that I have to use $u$ substitution but I can't seem to find something to substitute with.
100738
Calculation of $\displaystyle \int\sqrt{\tan x+2}\;dx$  $\bf{My\; solution::}$ Let $\displaystyle \tan x+2 = t^2 $, Then   $$\displaystyle \sec^2 (x)dx = 2tdt\Rightarrow dx = \frac{2t}{1+\tan^2 x}dt = \frac{2t}{1+(t^2-2)^2}dt$$  So Integral convert into $$\displaystyle \int \frac{2t^2}{(t^2-2)^2+1^2}dt$$  Let $\displaystyle (t^2-2) = u\Rightarrow t^2=u+2\;,$ Then $\displaystyle tdt=\frac{1}{2}du$  Now How can i solve after that  please help me  Thanks