What is the integral of t(t+1)^6 ?

The integral of t(t+1)^6 is (t^8)/8+(6t^7)/7+(5t^6)/2+4t^5+(15t^4)/4+2t^3+(t^2)/2+C

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integral of t(t+1)^6

\int t (t + 1)^{6} d t

\frac{t ^{8}}{8} + \frac{6 t ^{7}}{7} + \frac{5 t ^{6}}{2} + 4 t^{5} + \frac{15 t ^{4}}{4} + 2 t^{3} + \frac{t ^{2}}{2} + C

Solution steps

\int t (t + 1)^{6} d t

Expand

t (t + 1)^{6} : t^{7} + 6 t^{6} + 15 t^{5} + 20 t^{4} + 15 t^{3} + 6 t^{2} + t

= \int t^{7} + 6 t^{6} + 15 t^{5} + 20 t^{4} + 15 t^{3} + 6 t^{2} + t 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

= \int t^{7} d t + \int 6 t^{6} d t + \int 15 t^{5} d t + \int 20 t^{4} d t + \int 15 t^{3} d t + \int 6 t^{2} d t + \int t d t

\int t^{7} d t = \frac{t ^{8}}{8}

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

\int 15 t^{5} d t = \frac{5 t ^{6}}{2}

\int 20 t^{4} d t = 4 t^{5}

\int 15 t^{3} d t = \frac{15 t ^{4}}{4}

\int 6 t^{2} d t = 2 t^{3}

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

= \frac{t ^{8}}{8} + \frac{6 t ^{7}}{7} + \frac{5 t ^{6}}{2} + 4 t^{5} + \frac{15 t ^{4}}{4} + 2 t^{3} + \frac{t ^{2}}{2}

Add a constant to the solution

= \frac{t ^{8}}{8} + \frac{6 t ^{7}}{7} + \frac{5 t ^{6}}{2} + 4 t^{5} + \frac{15 t ^{4}}{4} + 2 t^{3} + \frac{t ^{2}}{2} + C

f (x) = x^{2} - 1, at x = 2

y = e^{x}, y = e^{2 x}, x = 0, x = ln (2)

\frac{x - ln ( 1 + x )}{x ^{2}}

f (x) = 4 x^{2} - 4 x + 1, at x = 0

\int 5^{2} - x^{2} d x

What is the integral of t(t+1)^6 ?
The integral of t(t+1)^6 is (t^8)/8+(6t^7)/7+(5t^6)/2+4t^5+(15t^4)/4+2t^3+(t^2)/2+C