Question

how to prove the following integral i think i know the way but i need some guidance through it please

Integral x(x+a)^n dx = (x+a)^n+1 ((n+1) x - a) / (n+1)(n+2)

Answer 1

this is how i tried to solve it

let u=x
u'=1dx

v'=(x+a)^n
v=integral (x+a)^n dx = (x+a)^n+1 / (n+1)

Answer 2

am i right so far?

Answer 3

yeah do u know integration by parts?

Answer 4

yeah could you please see if im getting it right please

Answer 5

= uv - integral u'v

Answer 6

Integral x(x+a)^n dx =x(x+a)^n+1 / (n+1) - integral 1 * (x+a)^n+1 / (n+1) dx

Answer 7

yeah but i think its not required here
\[take~~u=x+a\\then ~~dx=du\]
ok?

Answer 8

ok

Answer 9

and
also x=u-a
right

Answer 10

could you please show it to me step by step

Answer 11

ive got my notebook next to me

Answer 12

yeah i'll show u

Answer 13

thank you so much

Answer 14

now try to replace them in the integral it will be--
\[\Large \int\limits_{}^{}x(x+a)^ndx=\int\limits_{}^{}(u-a)*u^ndu\]
ok ?

Answer 15

how ?

Answer 16

what is u?

Answer 17

\[\because ~(x+a)=u\\and ~~then ~~x=(u-a)\\and ~also ~(x+a)^n=u^n\\OK?\]

Answer 18

understood

Answer 19

just used u-substitution

Answer 20

ok

Answer 21

is there a trick to know this?

Answer 22

nope u just have to think the possible ways to make it simpler
now multiplying this we have\[\Large \int\limits_{}^{}[u^{n+1}-au^n]dx=\int\limits_{}^{}u^{n+1}dx-\int\limits_{}^{}au^ndx\]
getting this?

Answer 23

let me write it down

Answer 24

sorry those dx will be du

Answer 25

i didnt get this part

Answer 26

\[\Large \int\limits_{}^{}u^{n+1}du-\int\limits_{}^{}au^n du\]

Answer 27

ok i do now

Answer 28

you multiplied

Answer 29

yeah

Answer 30

and from earlier assumption
dx=du
ok?

Answer 31

now just simple integration

Answer 32

using
\[\Huge \int\limits_{}^{}x^ndx=\frac{ x^{n+1} }{ n+1 }+C\]

Answer 33

how is dx = dy?

Answer 34

ok i get it sorry

Answer 35

because u = x+a

u'= 1 dx

Answer 36

look from the first step i took
\[x+a=u\\now ~just~differentiating~both~sides~we~have\\dx=da+du\\but ~da=differentiation ~of~a~constant=0\\so ~just \\dx=du\]

Answer 37

yeah thats correct :)

Answer 38

what shall i do next ?

Answer 39

add 1 to u^n+1

Answer 40

/ by n+2

Answer 41

i guess ill know how to do it from here

Answer 42

yeah next use\[\int\limits_{}^{}x^{n}dx=\frac{ x^{n+1} }{ n+1 }\]
so \[\large \int\limits_{}^{}u^{n+1}du=\frac{ u^{n+2} }{ n+2 }\]
ok?

Answer 43

does "a" go outside the integral?

Answer 44

yeah it will
\[\int\limits_{}^{}au^ndu=a \int\limits_{}^{}u^ndu=a*\frac{ u^{n+1} }{ n+1 }\]

Answer 45

and then ill have to plug in u

Answer 46

thank you so much for your time

Answer 47

no problem buddy :)

Answer 48

ill probably need help when putting it all together :)

Answer 49

sorry for being a pain in the retrice:)

Answer 50

lol ;)

Answer 51

i took (n+1)(n+2) as the common denominator

Answer 52

i meant retricehehehe

Answer 53

i guess this website doesnt allow you to use such words lol

Answer 54

i need help @sidsiddhartha

Answer 55

@sidsiddhartha if you could please help me on the last bit

Answer 56

yeah :)

Answer 57

i solved it thanks :)

Answer 58

that was proof number 7 of integral now im doing proof 8 which is

integral 1 / (1+x^2) dx = inverse tan x

Answer 59

i know that the differentiation of inverse tan x is 1 / (1+x^2)

Answer 60

but how can we find the integral of 1 / (1+x^2) dx using substitution or any other method

Answer 61

if we use u=1+x^2 then the derivative of u would have an x in it

Answer 62

ok u have to use this form--\[\int\limits_{}^{}\frac{ dx }{ x^2+a^2 }=\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a}+c\]

Answer 63

let me have a look at it for a moment

Answer 64

ok so it is the same form because 1^2 is 1

Answer 65

then--