Find the value of $\int\int\frac{1}{16x^2+16x+10} \,dx$.

Question

TuteeHUB · Accepted Answer

$(x+1/2) cos^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)$

TuteeHUB · Answer

$\frac{1}{8} (x+\frac{1}{2})Sin^{-1} (x+\frac{1}{2})-\frac{1}{2} ln⁡(1+(x+\frac{1}{2})^2)$

TuteeHUB · Answer

$\frac{1}{8} (x+\frac{1}{2})tan^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)$

TuteeHUB · Answer

$(x+1/2) cos^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)$

TuteeHUB · Answer

$\frac{1}{8} (x+\frac{1}{2}) sec^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)$

1.	Find the value of \(\int\int\frac{1}{16x^2+16x+10} \,dx\).
A.	\(\frac{1}{8} (x+\frac{1}{2})Sin^{-1} (x+\frac{1}{2})-\frac{1}{2} ln⁡(1+(x+\frac{1}{2})^2)\)
B.	\(\frac{1}{8} (x+\frac{1}{2})tan^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)
C.	\((x+1/2) cos^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)
D.	\(\frac{1}{8} (x+\frac{1}{2}) sec^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)
Answer» C. \((x+1/2) cos^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)

Find the value of \(\int\int\frac{1}{16x^2+16x+10} \,dx\).