What is the integral of ((x-1)^5+3(x-1)^2+5) ?

The integral of ((x-1)^5+3(x-1)^2+5) is 1/6 (x-1)^6+x^3-3x^2+8x+C

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integral of ((x-1)^5+3(x-1)^2+5)

\int ((x - 1)^{5} + 3 (x - 1)^{2} + 5) d x

\frac{1}{6} (x - 1)^{6} + x^{3} - 3 x^{2} + 8 x + C

Solution steps

\int (x - 1)^{5} + 3 (x - 1)^{2} + 5 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 (x - 1)^{5} d x + \int 3 (x - 1)^{2} d x + \int 5 d x

\int (x - 1)^{5} d x = \frac{( x - 1 ) ^{6}}{6}

\int 3 (x - 1)^{2} d x = x^{3} - 3 x^{2} + 3 x

\int 5 d x = 5 x

= \frac{( x - 1 ) ^{6}}{6} + x^{3} - 3 x^{2} + 3 x + 5 x

Simplify

\frac{( x - 1 ) ^{6}}{6} + x^{3} - 3 x^{2} + 3 x + 5 x : \frac{1}{6} (x - 1)^{6} + x^{3} - 3 x^{2} + 8 x

= \frac{1}{6} (x - 1)^{6} + x^{3} - 3 x^{2} + 8 x

Add a constant to the solution

= \frac{1}{6} (x - 1)^{6} + x^{3} - 3 x^{2} + 8 x + C

\frac{\partial}{\partial x} (e^{x} (sin (y) - y))

x \to + 1 + lim (2 x - 2 - 2)

\int \frac{1}{x - 7} d x

\int \frac{6 x - 5}{3 x ^{2} + 5} d x

y^{^{'}} = \frac{x ^{2} y ^{4} + 2 x ^{2}}{y ^{3}}

What is the integral of ((x-1)^5+3(x-1)^2+5) ?
The integral of ((x-1)^5+3(x-1)^2+5) is 1/6 (x-1)^6+x^3-3x^2+8x+C