What is the integral of 8/((4-t)^8) ?

The integral of 8/((4-t)^8) is 8/(7(4-t)^7)+C

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integral of 8/((4-t)^8)

\int \frac{8}{( 4 - t ) ^{8}} d t

\frac{8}{7 ( 4 - t ) ^{7}} + C

Solution steps

\int \frac{8}{( 4 - t ) ^{8}} d t

Take the constant out:

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

= 8 \cdot \int \frac{1}{( 4 - t ) ^{8}} d t

Apply u-substitution

: \int - \frac{1}{u ^{8}} d u

= 8 \cdot \int - \frac{1}{u ^{8}} d u

Take the constant out:

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

= 8 (- \int \frac{1}{u ^{8}} d u)

Apply the Power Rule

: - \frac{1}{7 u ^{7}}

= 8 (- (- \frac{1}{7 u ^{7}}))

Substitute back

u = 4 - t

= 8 (- (- \frac{1}{7 ( 4 - t ) ^{7}}))

Simplify

8 (- (- \frac{1}{7 ( 4 - t ) ^{7}})) : \frac{8}{7 ( 4 - t ) ^{7}}

= \frac{8}{7 ( 4 - t ) ^{7}}

Add a constant to the solution

= \frac{8}{7 ( 4 - t ) ^{7}} + C

y = 8 + 4 x^{2} - 2 x^{3}, (2, 8)

x \to 2 lim (x^{2} + 3 x - 4)

\int \frac{x}{x ^{4} + 7} d x

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

\int \frac{3 + 6 x}{1 + x ^{2}} d x

What is the integral of 8/((4-t)^8) ?
The integral of 8/((4-t)^8) is 8/(7(4-t)^7)+C