What is the integral of ((3inx-5)^4)/x ?

The integral of ((3inx-5)^4)/x is (81i^4n^4x^4)/4-180i^3n^3x^3+675i^2n^2x^2-1500inx+625ln|x|+C

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integral of ((3inx-5)^4)/x

\int \frac{( 3 in x - 5 ) ^{4}}{x} d x

\frac{81 i ^{4} n ^{4} x ^{4}}{4} - 180 i^{3} n^{3} x^{3} + 675 i^{2} n^{2} x^{2} - 1500 in x + 625 ln ∣ x ∣ + C

Solution steps

\int \frac{( 3 in x - 5 ) ^{4}}{x} d x

Expand

\frac{( 3 in x - 5 ) ^{4}}{x} : 81 i^{4} n^{4} x^{3} - 540 i^{3} n^{3} x^{2} + 1350 i^{2} n^{2} x - 1500 in + \frac{625}{x}

= \int 81 i^{4} n^{4} x^{3} - 540 i^{3} n^{3} x^{2} + 1350 i^{2} n^{2} x - 1500 in + \frac{625}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 81 i^{4} n^{4} x^{3} d x - \int 540 i^{3} n^{3} x^{2} d x + \int 1350 i^{2} n^{2} x d x - \int 1500 in d x + \int \frac{625}{x} d x

\int 81 i^{4} n^{4} x^{3} d x = \frac{81 i ^{4} n ^{4} x ^{4}}{4}

\int 540 i^{3} n^{3} x^{2} d x = 180 i^{3} n^{3} x^{3}

\int 1350 i^{2} n^{2} x d x = 675 i^{2} n^{2} x^{2}

\int 1500 in d x = 1500 in x

\int \frac{625}{x} d x = 625 ln ∣ x ∣

= \frac{81 i ^{4} n ^{4} x ^{4}}{4} - 180 i^{3} n^{3} x^{3} + 675 i^{2} n^{2} x^{2} - 1500 in x + 625 ln ∣ x ∣

Add a constant to the solution

= \frac{81 i ^{4} n ^{4} x ^{4}}{4} - 180 i^{3} n^{3} x^{3} + 675 i^{2} n^{2} x^{2} - 1500 in x + 625 ln ∣ x ∣ + C

\frac{d y}{d x} = 5 y^{2} - \frac{y}{x}

y = x^{2} - 1 arcsec (x)

6 (t) e^{- 2 t}

\frac{\partial}{\partial x} ((1 + 2 x y^{2})^{3})

y = \frac{t}{( t - 5 ) ^{2}}

What is the integral of ((3inx-5)^4)/x ?
The integral of ((3inx-5)^4)/x is (81i^4n^4x^4)/4-180i^3n^3x^3+675i^2n^2x^2-1500inx+625ln|x|+C