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

\int 7 cos^{2} (x t a) n^{3} x d x

n = 0 \sum \infty \frac{1}{4 n}

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

10 t^{\frac{6}{5}} + 6 e^{3}

\frac{d y}{d x} = y + 8

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