# f(x) = 1/x  + 6x^2     find the integral for f(x)

hala718 | High School Teacher | (Level 1) Educator Emeritus

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f(x) = 1/x  + 6x^2

intg f(x) = intg (1/x + 6x^2) dx

= intg (1/x)dx  + 6 intg (x^2)dx

We know that intg (1/x) = ln x

==>  intg f(x)  = ln x + 6 *x^3/3 + C

= ln x + 2x^3 + C

==> intg f(x) = ln x + 2x^3 + C

william1941 | College Teacher | (Level 3) Valedictorian

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We have to find the integral of f(x) where f(x) = 1/x  + 6x^2

Now for 1/x , Int [ 1/x ]= ln x

and for x^n , Int[ x^n] = x^(n+1)/ (n+1)

So here f(x) = 1/x  + 6x^2

=> Int [ 1/x  + 6x^2 ]

=> Int [1/x]  + Int [6x^2 ]

=> ln x + (6/3)*x^3

=> ln x  + 2 x^3

The required result is ln x  + 2 x^3

neela | High School Teacher | (Level 3) Valedictorian

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Let I =Int f(x) dx = Int (1/x+6x^2) dx

I = Int(1/x)dx +Int(6x^2)dx

I = lnx + 6* (1/(2+1)) x^(2+1), as int x^n dx = (1/n+1)x^(n+1).

I = lnx + 2x^3 + const

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The indefinite integral is:

Int f(x)dx = Int (1/x  + 6x^2)dx

We'll use the additive property of integral:

Int (1/x  + 6x^2)dx = Int (1/x)dx  + Int (6x^2)dx

Int (1/x)dx = ln |x| + C

Int (6x^2)dx = 6 Int x^2dx

6 Int x^2dx = 6x^3/3 + C

6 Int x^2dx = 2x^3 + C

Int (1/x  + 6x^2)dx = ln |x| + 2x^3 + C

Note: We've considered the absolute value of x, because it is not assigned an interval of values for x.