What is the integral of 7(t^5-t^3+1) ?

The integral of 7(t^5-t^3+1) is 7((t^6)/6-(t^4)/4+t)+C

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integral of 7(t^5-t^3+1)

\int 7 (t^{5} - t^{3} + 1) d t

7 (\frac{t ^{6}}{6} - \frac{t ^{4}}{4} + t) + C

Solution steps

\int 7 (t^{5} - t^{3} + 1) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int t^{5} - t^{3} + 1 d t

Apply the Sum Rule:

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

= 7 (\int t^{5} d t - \int t^{3} d t + \int 1 d t)

\int t^{5} d t = \frac{t ^{6}}{6}

\int t^{3} d t = \frac{t ^{4}}{4}

\int 1 d t = t

= 7 (\frac{t ^{6}}{6} - \frac{t ^{4}}{4} + t)

Add a constant to the solution

= 7 (\frac{t ^{6}}{6} - \frac{t ^{4}}{4} + t) + C

(- 3.3) (- 3.13)

\frac{x}{x ^{3}}

x^{2} - 2 x - 1, - 1, 2

\int x^{2} + y^{2} + z^{2} d x

\int \frac{6 x ^{3} - 8 x}{x ^{4} - 16} d x

What is the integral of 7(t^5-t^3+1) ?
The integral of 7(t^5-t^3+1) is 7((t^6)/6-(t^4)/4+t)+C