Find the indefinite integral below. $$\int \left( \frac { - 6 } { t ^ { 3 } } \mathbf { i } + 25 t ^ { 4 } \mathbf { j } + 4 t ^ { - 1 / 5 } \mathbf { k } \right) d t$$
Do not include an arbitrary constant vector.
A) $\frac { 3 } { t ^ { 2 } } \mathbf { i } + 5 t ^ { 5 } \mathbf { j } + 5 t ^ { 4 / 5 } \mathbf { k }$
B) $\frac { 3 } { t ^ { 2 } } \mathbf { i } + 5 t ^ { 5 } \mathbf { j } + 5 t ^ { 5 / 4 } \mathbf { k }$
C) $\frac { 3 } { t ^ { 3 } } \mathbf { i } + 5 t ^ { 5 } \mathbf { j } + 5 t ^ { 4 / 5 } \mathbf { k }$
D) $\frac { 3 } { t ^ { 2 } } \mathbf { i } + 5 t ^ { 5 } \mathbf { j } - 5 t ^ { 4 / 5 } \mathbf { k }$
E) $- \frac { 3 } { t ^ { 2 } } \mathbf { i } + 5 t ^ { 5 } \mathbf { j } + 5 t ^ { 4 / 5 } \mathbf { k }$

Question

Quizplus · Accepted Answer

Integrating each term: For $-\frac{6}{t^3}$, the integral is $\frac{3}{t^2}$ (using the power rule, increase the exponent by 1 and divide by the new exponent); for $25t^4$, the integral is $5t^5$ (again, power rule); for $4t^{-1/5}$, the integral is $5t^{4/5}$ (power rule).

Find the Indefinite Integral Below ∫(−6t3i+25t4j+4t−1/5k)dt\int \left( \frac { - 6 } { t ^ { 3 } } \mathbf { i } + 25 t ^ { 4 } \mathbf { j } + 4 t ^ { - 1 / 5 } \mathbf { k } \right) d t∫(t3−6​i+25t4j+4t−1/5k)dt

Find the Indefinite Integral Below $\int \left( \frac { - 6 } { t ^ { 3 } } \mathbf { i } + 25 t ^ { 4 } \mathbf { j } + 4 t ^ { - 1 / 5 } \mathbf { k } \right) d t$