How good are YOU ?
Here are some of my favourite integrals, ranging from beginners to just no.
Some of these have been posted before, most have not. Have fun.
Have fun and gl
Easier problems at the top, harder problems at the bottom. I really love all of these
Easy
I \, = \, \int {\frac{x}{e^x}} dx
I \, = \, \int_0^{\frac{\pi }{3n}} \tan \left( {nx} \right) dx
I \, = \, \int \frac{x^2+1}{x(x^2+3)} dx
I \, = \, 1+\int_{229}^{1234} (\sin(x)+\cos(x))^2+(\cos(x)-\sin(x))^2 \, dx
I \, = \, \int {\frac{1}{x\ln x}} dx
I \, = \, \int {\frac{{{e^x} + 1}}{{{e^x} - 1}}} dx
I \, = \, \int \frac{\sin(x)}{\sin(x)-1} dx
I \, = \, \int {2^{\ln(x)}} dx
I \, = \, \int {{e^{x + {e^x}}}} dx
I \, = \, \int_0^4 {\frac{\ln \left( x \right)}{\sqrt x}} dx
I \, = \, \int {{e^{\sqrt x }}} dx
I \, = \, \int \frac{1}{x^7-x} dx
I \, = \, \int\limits_0^4 {\frac{1}{{1 + \sqrt x }}} dx
I \, = \, \int \frac{\cos \left( x \right) - \sin \left( x \right)}{\sin \left( x \right) + \cos \left( x \right)} \, dx
I \, = \, \int \frac{\sin(x)\cos(x)}{\cos(x)^4-\sin(x)^4} \, dx
I \, = \, \int \frac{\sqrt{\sqrt{\ln(x)}+1}}{x} dx
I \, = \, \int \frac{1}{1+\sin\left( \frac{\pi}{6} \right)^x} dx
I \, = \, \int {\frac{1}{x \ln{{\left( x \right)}^n}}} dx
I \, = \, \int {\frac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} dx
I \, = \, \int \frac{2010}{x(1+x^{2010})} dx
Find the area between the function f(x) and the x-axis when f(x)=\sqrt{a-\sqrt{x}}
Show that \int \left( {x + 3} \right) \left( {x - 1} \right)^5 dx \: equals \: \frac{1}{21}\left( {3x + 11} \right){{\left( {x - 1} \right)}^6} + C
Medium
I \, = \, \int \sin(\ln(x)) + \cos(\ln(x)) dx
I \, = \, \int {\left( {1 + 2{x^2}} \right){e^{{x^2}}}} \, dx
I \, = \, \int \frac{\ln(x)-1}{\ln(x)^2} \, dx
I \, = \, \int_{\pi}^{3\pi} \sin(x)\ln(x) - \frac{\cos(x)}{x} \, dx
I \, = \, \int \frac{1}{\ln(x)}+\ln(\ln(x)) \, dx
I \, = \, \int (x+3) \sqrt{e^x \cdot x} dx
I \, = \, \int \arccos(x) + \arcsin(x) dx
I \, = \, \int \frac{1}{\cos(x)} dx
I \, = \, \int \frac{4(\ln x)^2+1}{(\ln x)^{\frac 32}}\ dx
I \, = \, \int \frac{1}{\sqrt[3]{x}+x}\, dx
I \, = \, \int \frac{\sin(x)+\sin(3x)}{\cos(3x)+\cos(x)} \, dx
I \, = \, \int \frac{1}{\sqrt{x+x\sqrt{x}}} \, dx
I \, = \, \int \sin(101x)\cdot\sin(99x) \, dx
I \, = \, \int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} \, dx
I \, = \, \int_0^1 \frac{x}{1+x}\sqrt{1-x^2}\ dx
I \, = \, \int \frac{1}{x^{a+1}+x} \, dx \; a>0
I \, = \, \int \frac{2^x \cdot 3^x}{9^x- 4^x} \, dx
I \, = \, \int \frac{1 + 2x^2}{x^5 \left( 1 + x^2 \right)^3} \, dx
I \, = \, \int \frac{1}{\sqrt{5x+3}-\sqrt{5x-2}} \, dx
I \, = \, \int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^2+1}}
I \, = \, \int_1^e \frac{\ln x-1}{x^2-(\ln x)^2}dx
I \, = \, \int_{1}^{\sqrt{3}} x^{2x^2+1}+\ln(x^{2x^{2x^2+1}}) dx
The integral \; \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sin {{\left( {2x} \right)}^3}\cos {{\left( {3x} \right)}^2} \, dx }
can be written as \; \left( \frac{a}{b} \right)^b where a and b are integers. Find \sqrt{a^b+b^a+1}
I \, = \, \int \frac{\left( x + 2 \right)^5}{\left( x + 7 \right)^2} \, dx
I \, = \, \int \frac{\sinh(x)+\cosh(x)}{\cosh(x)-\sinh(x)} \, dx
I \, = \, \int\limits_0^{\pi /2} {\sqrt {1 - 2\sin \left( {2x} \right) + 3\cos {{\left( x \right)}^2}} dx}
I \, = \, \int_{0}^{\pi/2} \ln(sec(x))
I \, = \, \int_{0}^{a} \frac{1}{\sqrt{x^2+a^2}} dx
I \, = \, \int \sqrt{ \sqrt{x+2\sqrt{2x+4}} + \sqrt{x-2\sqrt{2x+4}} } dx
I \, = \, \int \frac{ \left( 1 + x^2 \right) } { \left( 1 - x^2 \right) \sqrt{1 + x^4} } \, dx
I \, = \, \int \frac{ \sin(x) + \cos(x) }{e^x + 3 \cos(x) } \, dx
I \, = \, \int \sqrt{ \frac{k + x}{x} } \, dx
I \, = \, \int_{0}^{\infty} x^n \cdot e^{-x} \, dx
I \, = \, \int {\frac{x}{1+\cos(x)}} \, dx
I \, = \, \int {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \, dx
I \, = \, \int\limits_{ - \pi }^\pi {\sqrt {1 + \cos \left( x \right)} } \, dx
I \, = \, \int e^{x/2} \cdot \left( \frac{2 - \sin \left( x \right) }{1 - \cos \left( x\right) } \right) \, dx
I \, = \, \int_{0}^{\infty} \frac{\{1-(x-1)e^x\}\ln(x)}{(1+e^x)^2} dx
I \, = \, \int\limits_0^{\ln \left( 2 \right)} {\sqrt {\frac{{{e^x} + 1}}{{{e^x} - 1}}} } \, dx
I \, = \, \int_{0}^{\infty} \frac{1}{ \left( x + \sqrt{1+x^2} \right)^{\phi}} \, dx \qquad \phi=\frac{1+\sqrt{5}}{2}
I \, = \, \int_0^{\infty} \frac{1}{(x^2+2)^3} \, dx
I \, = \, \int \sqrt{\frac{1}{\,\sin(x)+1}\,}\, dx
I \, = \, \int {\sqrt[3] {\tan \left( x \right)} } \, dx
I \, = \, \int_{0}^{\infty} \frac{x^{29}}{(5x^2+49)^{17}}\, dx \, = \, \frac{14!}{2 \cdot 5^{15} \cdot 49^2 \cdot 16! }\, dx
I \, = \, \int \left( \frac{\arctan(x)}{x-\arctan(x)} \right)^2 \, dx
"Fun"
I \, = \, \int_{1}^{\infty} \ln\left( \sqrt[X]{x} \right)^{2011} dx
I \, = \, \int_{0}^{2\pi} \frac{x \sin(x)}{1+\sin(x)^2} dx
I \, = \, \int_{0}^{2\pi} {\frac{1}{1+e^{\cos(x)}}} dx
I \, = \, \int_0^{\frac {\pi}{2}} \frac{(\sin x)^{\cos x}}{(\cos x)^{\sin x} + (\sin x)^{\cos x}} dx
I \, = \, \int_2^{4} \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}} dx
I \, = \, \int_{-1}^{1} \arctan(e^x) dx
I \, = \, \int_{0}^{1} \ln(x)\cdot\ln(1-x) dx
I \, = \, \int_{-\pi/2}^{\pi/2} \frac{1}{1+2009^x} \cdot \frac{\sin(2010x)}{\sin(2010x)+\cos(2010x)} dx
I \, = \, \int \frac{1}{(x^2+1)^k} dx
I \, = \, \int_{0}^{\infty} \frac{x}{e^x-1} dx
I \, = \, \int_{0}^{a} \frac{1}{x+\sqrt{a^2-x^2}} dx \quad a>0
I \, = \, \int_{0}^{\infty} \frac{\ln{2x}}{x^2+9} dx
I \, = \, \int_{0}^{\pi/2} \ln(1-a \cos(x)) \, dx
I \, = \, \int_{-\infty}^{\infty} \sin(x^2)+\cos(x^2) dx
I \, = \, \int_{-\infty}^{\infty} \frac{x^a+x^b}{\ln(x)} dx \qquad (a,b)\forall >-1
I \, = \, \int_0^1 \frac{\ln(1+x^2)}{1+x^2} dx
I \, = \, \int_{0}^{\infty}\frac{1}{x^n+1} dx \quad n>1
I \, = \, \int_{0}^{\pi} \frac{1-\cos{(n\cdot x)}}{1-\cos(x)} dx \quad n\in\mathbb{N^+}
I \, = \, \int_{0}^{1} \ln(x) \cdot e^{-x} \cdot (1-x) dx
I \, = \, \int_{0}^{\infty} \ln\left( \frac{e^x+1}{e^x-1}\right) dx
I \, = \, \int_{0}^{\pi} \ln \left( 1 - 2\alpha \cos(x) + \alpha^2 \right) dx
I \, = \, \int_0^{\infty} \ln (1+e^{-ax} ) dx \quad a>0
I \, = \, \int_0^{\infty} 1 - x\sin\left( \frac{1}{x} \right) dx
I \, = \, \int_0^{\infty} \ln (1+e^{-ax} ) dx \quad a>0
I = \int_{0}^{\infty} \frac{{{e^{ - x}}\left( {1 - {e^{ - 6x}}} \right)}}{{x\left( {1 + {e^{ - 2x}} + {e^{ - 4x}} + {e^{ - 6x}} + {e^{ - 8x}}} \right)}}dx
I = \int\limits_0^1 {\frac{{\sin \left( {p\ln x} \right) \cdot \cos \left( {q\ln x} \right)}}{{\ln x}}}dx
I \, = \, \int_{a}^{\infty} \frac{1}{x^{n+1} \cdot \sqrt{x^2-a^2}} \, dx n \in \mathbb{N}^{+} \; , \; a>0
I \, = \, \int_{a}^{\infty} \frac{x^{-p}}{1+x} \, dx
I \, = \, \int_{ - \infty }^{\infty} \frac{x\sin x}{1 + x^2}dx
I \, = \, \int_0^{\infty} \sin(x)\cdot\arctan\left( \frac{1}{x} \right) dx
I \, = \, \int_{0 }^{\infty} \frac{1}{x} \cdot \sin(\tan(x)) \, dx
Have fun and gl.