Question

help with indefinite integral

Answer 1

it's not (9t^2-78t+320log(3t+13)-507)/9

Answer 2

First of all you can simplify it into a proper fraction as the degree of polynomial in numerator = 2> degree of polynomial in denominator (i.e. 1)

Answer 3

What will you get ?

Answer 4

6t-6/16

Answer 5

No no not like that.

Answer 6

See:

\[6t^2 - 6 = 2t(3t + 13) - 26t - 6\]

Answer 7

So you integral would become:

\[\int \cfrac{2t(3t+13)-26t-6}{3t+13}dt\]

\[= \int(2t)dt - \int\cfrac{26t+6}{3t+13}dt\]

Answer 8

Now can you evaluate the first integral ?

Answer 9

yeah its 2

Answer 10

Nope

Answer 11

i mean t^2

Answer 12

Yup great

Answer 13

plus C

Answer 14

We will write plus C after evaluating both the integrals

Answer 15

Okay now lets see if we can evaluate the second integral.

Answer 16

Any idea ?

Answer 17

sup rule so int 26t/(3t+13) + int 6/3t+13

Answer 18

(26t/3)-(320ln(3t+13))/9

Answer 19

@vishweshshrimali5

Answer 20

Sorry internet problem

Answer 21

its all good

Answer 22

okay lets see, your first step was perfect. Now I just have to verify your last step :)

Answer 23

okay how did you obtain the last step ?

Answer 24

its not a step but the answer to the second integral

Answer 25

Okay let me check.

Answer 26

Okay are you sure you are not forgetting some number like 338/9?

Answer 27

Or have you included it in C

Answer 28

well i don't think i did it correctly then because i don't remember using 338/9

Answer 29

Okay lets see:

\[\large{\int\cfrac{26t+6}{3t+13}dt}\]

\[\large{= \int \cfrac{26t}{3t+13}dt + 6\int \cfrac{1}{3t+13}dt}\]

\[\large{= I_2 + I_3}\]

Answer 30

ok

Answer 31

\[\large{I_2 = 26\int\cfrac{t}{3t+13}dt}\]

Put, \(\large{u = 3t+13}\)

\[\large{\implies du = 3dt}\]

Thus, we have,

\[\large{I_2 = 26\int\cfrac{\cfrac{u-13}{3}}{u}\cfrac{du}{3}}\]

\[\large{= \cfrac{26}{9}\int\cfrac{u-13}{u}du}\]

\[\large{= \cfrac{26}{9}[u - 13\ln u] + C}\]

Got this ?

Answer 32

\[\large{= \cfrac{26}{9}[3t+13-13\ln(3t+13)]}\]

Okay ?

Answer 33

taking notes.. yeah this is helping

Answer 34

Good

Answer 35

Now \(\large{I_3}\) is easier to solve. Lets see:

\[\large{I_3 = 6\int\cfrac{1}{3t+13}dt}\]

As in \(\large{I_2}\), we substitute,

\[\large{u = 3t+13}\]

\[\large{\implies du = 3dt}\]

So, we have,

\[\large{I_3 = 6\int\cfrac{1}{u}\cfrac{du}{3}}\]

\[\large{= 2\ln{u} + C'}\]


Any doubt ?

Answer 36

need a second to catch up

Answer 37

take ur time

Answer 38

ok

Answer 39

gud

Answer 40

now again we replace u by 3t+13

Answer 41

2ln(3t+13)+C'

Answer 42

gr8

Answer 43

so I2 + I3 becomes ?

Answer 44

(26t/3)-(320/9)ln(3t+13)+(338/9)

Answer 45

fantastic !

Answer 46

So our final answer becomes:

t^2 + I_2 + I_3

Answer 47

muchas gracias!

Answer 48

:) my pleasure

Answer 49

You too did a great work

Answer 50

thanks