Question

Solve the equation on a piece of paper and post the answer as an attachment! ​

Answer 1

I=∫(x(π+49))157π2(xπ+7)dx

We can use π≈227 to avoid using the Hypergeometric Functions,

so substitute

u=xπ+7

⟹du=πxπ−1dx

⟹du=πx227−1dx

⟹du=πx227−77dx

⟹du=πx22–77dx

⟹du=πx157dx

⟹dx=duπx157

I=∫(x(π+49))157π2uduπx157

=∫x157(π+49)157π2uduπx157

=(π+49)157π3∫1udu

=(π+49)157ln(u)π3

=(π+49)157ln(xπ+7)π3

=(π+49)π−1ln(xπ+7)π3+c

Answer 2

hope you'll understand