What is the integral of 4x^3-1/(1-x) ?

The integral of 4x^3-1/(1-x) is x^4+ln|1-x|+C

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integral of 4x^3-1/(1-x)

\int 4 x^{3} - \frac{1}{1 - x} d x

x^{4} + ln ∣ 1 - x ∣ + C

Solution steps

\int 4 x^{3} - \frac{1}{1 - 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 4 x^{3} d x - \int \frac{1}{1 - x} d x

\int 4 x^{3} d x = x^{4}

\int \frac{1}{1 - x} d x = - ln ∣ 1 - x ∣

= x^{4} - (- ln ∣ 1 - x ∣)

Simplify

= x^{4} + ln ∣ 1 - x ∣

Add a constant to the solution

= x^{4} + ln ∣ 1 - x ∣ + C

\int \frac{x}{( 4 - x ^{2} ) ^{\frac{3}{2}}} d x

\int 5 x^{\frac{1}{4}} - 7 x^{\frac{3}{4}} d x

f (x) = \frac{7}{x}

\int \frac{3 x ^{4}}{x ^{2} - 2 x} d x

\frac{- x ^{2} + x + 6}{e ^{x}}

What is the integral of 4x^3-1/(1-x) ?
The integral of 4x^3-1/(1-x) is x^4+ln|1-x|+C