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Greetings,

∫ 3x^2 + 2x - 1/x^2 dx

integrate term by term

x^3 + x^2 + 1/x + C

or was the integral

∫ (3x^2 + 2x - 1)/x^2 dx

∫ 3 + 2/x - 1/x^2 dx

3x + 2lnx + 1/x + C

Regards

Hi ,

Frankly speaking the the qn is one of the simpler qns i hve seen in calculus.

solve it as follows:

intergral([3x^2 + 2x - 1]/x^2 dx)

split the terms first

= integral(3) + integral(2/x) - intergral(1/X^2)

now in second term cancel the x in numerator and denominator and take the x from deno to num. changing the power to -1.

=x + integral(2x^-1) - integral(x^-2)

= x + ln(x) - x^-1.

= x + ln(x) - 1/x + c where c = constant

Theres a standard formula for integral(1/x) which is ln(x) + c

This is due to fact that derivative of ln(x) = 1/x and derivative is the inverse operator for intergral.

I took calculus about 4 yrs back. So cant say there wud be no mistakes ;)

njoy calculus.

actually this is pretty straight forward integration, no need for special formulas

∫ 3x^2 + 2x - 1/x^2 dx = ∫ 3x² + 2x - x^(-2)

= (3/3)x^3 + (2/2)x² - (1/-1)x^(-1)

= x^3 + x² + x^-1 or x^3 + x² + 1/x

the last term switched signs because for the antiderivative you add one to the exponent and divide by that sum, so -x / -1 = + x^-1

note: 1/x² = x^(-2)

∫ 3x^2 + 2x - 1/x^2 dx

= ∫ 3x^2 dx+ ∫2x dx- ∫ x^(-2) dx

for ∫ 3x^2 dx use ∫ax^n dx= a ((x^(n+1)) / (n+1))

substitute x^2 in x^n to yield and 3 for a

= 3(x^(2+1)) / (2+1) = 3(x^3) / 3 = x^3

for ∫2x dx use again ∫ax^n dx= a ((x^(n+1)) / (n+1))

to yield x^2

∫ x^(-2) dx = (x^(-2+1)) / (-2+1) = (x^(-1)) / -1 = - 1 / x

combine

∫ (3x^2 + 2x - 1/x^2) dx = x^3 + x^2 - ( -1/x)

= x^3 + x^2 +1/x + C

---------------------------------

ok, so you've revised the problem

first, simplify the inner function ∫ (3x^2 + 2x - 1) / (x^2) dx

= ∫ (3 + (2/x) - (1/x^2) )dx

= 3x + 2 ln x + 1/x + C