-1-5x-10x2-10x3-5x4-x5dx -1-xp-6-c then p is

The correct option is C 6 Given: ∫ ( 1+5x+10x^2+10x^3+5x^4+x^5 )dx. = ∫ ( x+1 )^5dx #160;[by binomial theorem] = ( x+1 )^5.dx= ( ...

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Standard XII

Mathematics

Properties of Iota

∫1+5x+10x2+10...

Question

$\int (1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5}) d x$ $= \frac{(1 + x)^{p}}{6} + c$ then $p$ is

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Solution

The correct option is C $6$
Given: $\int (1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5}) d x .$
$=$ $\int {(x + 1)}^{5} d x$ [by binomial theorem]
$= {(x + 1)}^{5} . d x = \frac{{(x + 1)}^{6}}{6} + c$
So, the value of $p = 6$

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∫(1+5x+10x2+10x3+5x4+x5)dx =(1+x)p6+c then p is

The correct option is C 6Given: ∫(1+5x+10x2+10x3+5x4+x5)dx.= ∫(x+1)5dx [by binomial theorem]=(x+1)5.dx=(x+1)66+cSo, the value of p=6

$\int (1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5}) d x$ $= \frac{(1 + x)^{p}}{6} + c$ then $p$ is

The correct option is C $6$
Given: $\int (1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5}) d x .$
$=$ $\int {(x + 1)}^{5} d x$ [by binomial theorem]
$= {(x + 1)}^{5} . d x = \frac{{(x + 1)}^{6}}{6} + c$
So, the value of $p = 6$