What is the integral of x^6-4x^4+x+1 ?

The integral of x^6-4x^4+x+1 is (x^7)/7-(4x^5)/5+(x^2)/2+x+C

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integral of x^6-4x^4+x+1

\int x^{6} - 4 x^{4} + x + 1 d x

\frac{x ^{7}}{7} - \frac{4 x ^{5}}{5} + \frac{x ^{2}}{2} + x + C

Solution steps

\int x^{6} - 4 x^{4} + x + 1 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^{6} d x - \int 4 x^{4} d x + \int x d x + \int 1 d x

\int x^{6} d x = \frac{x ^{7}}{7}

\int 4 x^{4} d x = \frac{4 x ^{5}}{5}

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

= \frac{x ^{7}}{7} - \frac{4 x ^{5}}{5} + \frac{x ^{2}}{2} + x

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{4 x ^{5}}{5} + \frac{x ^{2}}{2} + x + C

\int \frac{t ^{5}}{t ^{2} + 10} d t

\frac{d y}{d x} = 0.5 (100 - y)

\int_{1}^{5} x ln (x) d x

f (x) = x (x + 3)^{2}

\int_{4}^{\infty} \frac{1}{x ^{3}} d x

What is the integral of x^6-4x^4+x+1 ?
The integral of x^6-4x^4+x+1 is (x^7)/7-(4x^5)/5+(x^2)/2+x+C