Question

Integration by partial fractions:

Answer 1

\[\LARGE \int \frac{x^3+4}{x^2+4}~dx\]

Answer 2

\[\LARGE \int x~dx-\int\frac{4x-4}{x^2+4}~dx\]
\[\LARGE \int x~dx-4\int\frac{x-1}{x^2+4}~dx\]

Answer 3

Looks good, you may break the second integral into two -
1) work one integral using u-substitution
2) work the other integral using trig-substitution

Answer 4

Alright, just wasn't sure if I did it right up to that point, thank you :)

Answer 5

yesh its right! :)

Answer 6

absolutely right ! you did the long division correctly to reduce the degree of numerator and everything else looks right !

Answer 7

\[ \int x~dx-4\int\frac{x-1}{x^2+4}~dx = \int x~dx-4\int\frac{x}{x^2+4}~dx - 4\int\frac{1}{x^2+4}~dx \]