Using the result,

For $I(\lambda ) = \int\limits_a^b {f(x,\lambda )dx} $, $\frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_a^b {\left[ {\frac{\partial }{{\partial \lambda }}f(x,\lambda )} \right]dx} $

Let, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } – 1}}{{\ln x}}dx} $

$\frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_0^1 {\left[ {\frac{\partial }{{\partial \lambda }}\left( {\frac{{{x^\lambda } – 1}}{{\ln x}}} \right)} \right]dx} $ Continue reading $\int\limits_0^1 {\frac{{{x^{e – 1}} – 1}}{{\ln x}}dx} = ?$ →

The given integral =$ – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{x(1 + x)}}dx} $

$ = \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{1 + x}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = \int\limits_{ – 1}^0 {\frac{{\ln \left[ {1 + ( – 1 – x) + {{( – 1 – x)}^2}} \right]}}{{1 + (-1-x)}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{ – x}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = – 2\int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = – 2\int\limits_{ – 1}^0 {\frac{{\ln \frac{{(1 – x)(1 + x + {x^2})}}{{(1 – x)}}}}{x}dx} $ Continue reading Prove that,

$\int\limits_0^1 {\frac{1}{{x(1 + x)}}.\ln \left[ {\frac{1}{{1 + x(1 + x)}}} \right]dx} = $

$\frac{4}{3}\left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ……..\infty \, terms} \right)$ →

We have, $f(x) = – \frac{1}{{{x^2}}}f\left( {\frac{1}{x}} \right)$

$\therefore I = \int\limits_{\tan \alpha }^{\cot \alpha } { – \frac{1}{{{x^2}}}f\left( {\frac{1}{x}} \right)dx} $

Let, $\frac{1}{x} = t$

$ \Rightarrow – \frac{1}{{{x^2}}}dx = dt$

$\therefore I = \int\limits_{\cot \alpha }^{\tan \alpha } {f(t)dt} =-I $

$ \Rightarrow I = 0$

The given integral can be split as,

$ \int\limits_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 – x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}} dx + \int\limits_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{1 + \cos 2x}}} dx$

$\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 – x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}$ is an odd function. Thus, the given integral

$ = 0 + \int\limits_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{2{{\cos }^2}x}}} dx = \frac{1}{2}\int\limits_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} {{{\sec }^2}xdx} = \frac{1}{2}\left. {\tan x} \right|_{ – \frac{\pi }{6}}^{\frac{\pi }{6}} = \frac{1}{{\sqrt 3 }}$

Let $f:R \to R$ be a continuous function such that $f(x) + f(x + 1) = 2$, for all $x\in R$. If ${I_1} = \int\limits_0^8 {f(x)dx} $ and ${I_2} = \int\limits_{ – 1}^3 {f(x)dx} $, then the value of $I_1 +2I_2 $ is equal to …… Continue reading $f(x) + f(x + 1) = 2$

$\int\limits_0^8 {f(x)dx + 2\int\limits_{ – 1}^3 {f(x)dx} } = ?$ →