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

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

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integral of (1+(x+1)^4)/(x+1)

\int \frac{1 + ( x + 1 ) ^{4}}{x + 1} d x

ln ∣ x + 1 ∣ + \frac{( x + 1 ) ^{4}}{4} + C

Solution steps

\int \frac{1 + ( x + 1 ) ^{4}}{x + 1} d x

Expand

\frac{1 + ( x + 1 ) ^{4}}{x + 1} : \frac{1}{x + 1} + \frac{( x + 1 ) ^{4}}{x + 1}

Apply the Sum Rule:

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

= \int \frac{1}{x + 1} d x + \int \frac{( x + 1 ) ^{4}}{x + 1} d x

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

\int \frac{( x + 1 ) ^{4}}{x + 1} d x = \frac{( x + 1 ) ^{4}}{4}

= ln ∣ x + 1 ∣ + \frac{( x + 1 ) ^{4}}{4}

Add a constant to the solution

= ln ∣ x + 1 ∣ + \frac{( x + 1 ) ^{4}}{4} + C

\frac{8 x}{9 - cot ( x )}

\frac{d}{d x} (\frac{1}{1 - 2 x ^{2}})

y = 6 x - x^{2}

\int e^{t} t d t

y^{^{'}} sin (x) + y cos (x) = x sin (x)

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