What is the integral of ((m-6)x+(m+16))/(7x^2+3x-4) ?

The integral of ((m-6)x+(m+16))/(7x^2+3x-4) is

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integral of ((m-6)x+(m+16))/(7x^2+3x-4)

\int \frac{( m - 6 ) x + ( m + 16 )}{7 x ^{2} + 3 x - 4} d x

(x (m - 6) + m + 16) (\frac{1}{11} ln ∣ 7 x - 4 ∣ - \frac{1}{11} ln ∣ x + 1 ∣) - \frac{m - 6}{77} (7 x ln ∣ 7 x - 4 ∣ - 7 x ln ∣ x + 1 ∣ - 4 ln ∣ 7 x - 4 ∣ - 7 ln ∣ x + 1 ∣ + 11) + C

Solution steps

\int \frac{( m - 6 ) x + m + 16}{7 x ^{2} + 3 x - 4} d x

Apply Integration By Parts

: (x (m - 6) + m + 16) (\frac{1}{11} ln ∣ 7 x - 4 ∣ - \frac{1}{11} ln ∣ x + 1 ∣) - \int \frac{m - 6}{11} (ln ∣ 7 x - 4 ∣ - ln ∣ x + 1 ∣) d x

= (x (m - 6) + m + 16) (\frac{1}{11} ln ∣ 7 x - 4 ∣ - \frac{1}{11} ln ∣ x + 1 ∣) - \int \frac{m - 6}{11} (ln ∣ 7 x - 4 ∣ - ln ∣ x + 1 ∣) d x

\int \frac{m - 6}{11} (ln ∣ 7 x - 4 ∣ - ln ∣ x + 1 ∣) d x = \frac{m - 6}{77} (7 x ln ∣ 7 x - 4 ∣ - 7 x ln ∣ x + 1 ∣ - 4 ln ∣ 7 x - 4 ∣ - 7 ln ∣ x + 1 ∣ + 11)

= (x (m - 6) + m + 16) (\frac{1}{11} ln ∣ 7 x - 4 ∣ - \frac{1}{11} ln ∣ x + 1 ∣) - \frac{m - 6}{77} (7 x ln ∣ 7 x - 4 ∣ - 7 x ln ∣ x + 1 ∣ - 4 ln ∣ 7 x - 4 ∣ - 7 ln ∣ x + 1 ∣ + 11)

Add a constant to the solution

= (x (m - 6) + m + 16) (\frac{1}{11} ln ∣ 7 x - 4 ∣ - \frac{1}{11} ln ∣ x + 1 ∣) - \frac{m - 6}{77} (7 x ln ∣ 7 x - 4 ∣ - 7 x ln ∣ x + 1 ∣ - 4 ln ∣ 7 x - 4 ∣ - 7 ln ∣ x + 1 ∣ + 11) + C

\int sin (lo g_{e} (x)) d x

\int \frac{sin ( 3 x )}{3 + cos ( 3 x )} d x

6 x

\int (2 - 3 x)^{10} d x

\int \frac{tan ^{2} ( 2 x )}{sec ( 2 x )} d x

What is the integral of ((m-6)x+(m+16))/(7x^2+3x-4) ?
The integral of ((m-6)x+(m+16))/(7x^2+3x-4) is