r3 +3x +412.

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Solution

The integral is,

y= ∫ x 3 +3x+4 x dx

Here, y is the solution of integral.

Use the formula of ∫ x n dx = x n+1 n+1 +A, where Ais constant.

y= ∫ x 3 +3x+4 x dx = ∫ ( x 3 x +3 x x + 4 x ) dx = ∫ ( x 3− 1 2 +3 x 1− 1 2 +4 x − 1 2 ) dx = ∫ ( x 5 2 ) dx+3 ∫ ( x 1 2 ) dx+4 ∫ ( x −1 2 ) dx

Simplify further.

y= x 5 2 +1 5 2 +1 + 3 x 1 2 +1 1 2 +1 +4 x − 1 2 +1 − 1 2 +1 +D = x 7 2 7 2 +3 x 3 2 3 2 +4 x 1 2 1 2 +D = 2 x 7 2 7 +2 x 3 2 +8 x 1 2 +D

Where, D is constant.

Thus, the solution of integral is 2 x 7 2 7 +2 x 3 2 +8 x 1 2 +D.

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