How do you find the indefinite integral of #int 1/4(x)(7 + 6x^2)dx#?

Question

Isaiah Allen · Accepted Answer

I would factor out the #1/4# then distribute #x# and integrate term by term.

#int 1/4(x)(7 + 6x^2)dx = 1/4 int (7x+6x^3)dx# # = 1/4[(7x^2)/2+(6x^4)/4] +C# The answer may be rewritten to taste. Method 2 Do not factor out, but distribute the #1/4# #int 1/4(x)(7 + 6x^2)dx = int ((7x)/4+(6x^3)/4)dx# # = (7x^2)/8 + (6x^4)/16 +C# Method 3 Use substitution. #int 1/4(x)(7 + 6x^2)dx = 1/4 int (7+6x^2) x dx# Let #u = 7+6x^2#, so that #du = 12x dx# and #xdx = (du)/12# The integral becomes: #1/4 int u (du)/12 = 1/48 int u du# # = 1/48 u^2/2 +C# # = 1/96(7+6x^2)^2 +C# This method has an additional constant of #49/96# in it that is "absorbed" into the arbitrary constant #C# in the other solutions.

Scarlett Allison · Answer

undefined To find the indefinite integral of $ \int \frac{1}{4}x(7 + 6x^2) \, dx $, you can distribute the $ \frac{1}{4}x $ across the terms in the parentheses and then integrate each term separately.

$$
\int \frac{1}{4}x(7 + 6x^2) \, dx = \frac{1}{4} \left( \int 7x \, dx + \int 6x^3 \, dx \right)
$$

Now integrate each term:

$$
\int 7x \, dx = \frac{7}{2}x^2 + C_1
$$

$$
\int 6x^3 \, dx = \frac{6}{4}x^4 + C_2
$$

Finally, combine the results:

$$
\frac{1}{4} \left( \frac{7}{2}x^2 + C_1 + \frac{6}{4}x^4 + C_2 \right) = \frac{7}{8}x^2 + \frac{3}{8}x^4 + C
$$

where $ C $ is the constant of integration.