Let I=∫01x(1−x)n dx\int_{0}^{1} x(1-x)^n \,dx∫01x(1−x)ndx
∴I=∫01(1−x)(1−(1−x))n dx\int_{0}^{1} (1-x)(1-(1-x))^n \,dx∫01(1−x)(1−(1−x))ndx
=∫01x(1−x)(x)n dx\int_{0}^{1} x(1-x) (x)^n \,dx∫01x(1−x)(x)ndx
=∫01(xn−xn+1)dx=∫^1_0(x^n-x^{n+1})dx=∫01(xn−xn+1)dx
=[xn+1n+1−xn+2n+2] (∫0aƒ(x)dx=∫0aƒ(a−x)dx)=[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}]\,\,\,\,\, (∫^a_0ƒ(x)dx=∫^a_0ƒ(a-x)dx)=[n+1xn+1−n+2xn+2](∫0aƒ(x)dx=∫0aƒ(a−x)dx)
=[1n+1−1n+2][\frac{1}{n+1}-\frac{1}{n+2}][n+11−n+21]
=(n+2)−(n+1)(n+1)(n+2)\frac{(n+2)-(n+1)}{(n+1)(n+2)}(n+1)(n+2)(n+2)−(n+1)
=1(n+1)(n+2)\frac{1}{(n+1)(n+2)}(n+1)(n+2)1
If ∫ex(x3+x2−x+4) dx=exf(x)+C, \int e^x (x^3 + x^2 - x + 4) \, dx = e^x f(x) + C, ∫ex(x3+x2−x+4)dx=exf(x)+C, then f(1) f(1) f(1) is:
The value of : ∫x+1x(1+xex)dx \int \frac{x + 1}{x(1 + xe^x)} dx ∫x(1+xex)x+1dx.