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int8x(1)/(2x+1)dx

8x12x+1dx \int 8 x \frac{1}{2 x+1} d x

Full solution

Q. 8x12x+1dx \int 8 x \frac{1}{2 x+1} d x
  1. Define u-substitution: Let's do a u-substitution where u=2x+1u = 2x + 1, then du=2dxdu = 2dx.
  2. Rewrite in terms of uu: Rewrite the integral in terms of uu: 8x2x+1dx=4udu\int \frac{8x}{2x+1}\,dx = \int \frac{4}{u} \,du (since 8x=4(2x)8x = 4(2x) and 2dx=du2dx = du).
  3. Integrate 4/u4/u: Now, integrate 4u\frac{4}{u} with respect to uu: 4udu=4lnu+C\int \frac{4}{u} \, du = 4\ln|u| + C.
  4. Substitute back for u: Substitute back for u to get the integral in terms of x: 4ln2x+1+C4\ln|2x+1| + C.

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