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The integral value of
\int \left ( 4x+1 \right )^3dx

Option: 1

\frac{\left (4x-1 \right )^{4}}{16}+c


Option: 2

\frac{\left ( 4x+1 \right )^{4} }{16}+c


Option: 3

\frac{\left (-4x-1 \right )^{4}}{16}+c


Option: 4

\frac{\left (-4x+1 \right )^{4}}{16}+c


Answers (1)

best_answer

Given integral,
\int \left ( 4x+1 \right )^3dx
Rearranging the terms,
\int \left ( 4x+1 \right )^3dx=\frac{1}{4}\int \left ( 4x+1 \right )^3\cdot 4\ dx
Using the substitution method,
Let’s take u=4x+1
du=4\ dx
Thus,
\frac{1}{4}\int \left ( u \right )^3\cdot du
\frac{1}{4}\int \left ( u \right )^{3}\cdot du=\frac{1}{4}\cdot \frac{u^{4}}{4}+c
Substituting back the value of u=4x+1

Therefore,
\frac{1}{4}\int \left ( 4x+1 \right )^3\cdot 4\ dx=\cdot \frac{\left (4x+1 \right )^{4}}{16}+c

Posted by

Ritika Jonwal

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