Evaluate - - x 2 - 1 d x x 4 - 7 x 2 - 1

x^2+1x^4+7 x^2+1=1+1x^2x^2+1x^2+7=1+1x^2(x 1x)^2+(9) Put t=x 1x d t =(1+1x^2) d x ∫x^2+1x^4 7x^2+1d x=∫d tt^2+9=13tan^ 1(t3)+C=13tan^ 1(x 1x3)+C ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Special Integrals - 1

Evaluate : ...

Question

Evaluate : $\int \frac{(x^{2} + 1) d x}{x^{4} + 7 x^{2} + 1}$

Open in App

Solution

Suggest Corrections

Similar questions

\int \frac{(2 x + 3) (x^{2} - 3 x + 1) d x}{(x^{4} - 7 x^{2} + 1) (1 + ln (1 + 3 x + x^{2}))} = f (x) + C,

C

f (0) = 0,

\int_{0}^{\infty} \frac{x^{2} + 1}{x^{4} + 7 x^{2} + 1} d x

\int \frac{x^{2} + 1}{x^{4} + 7 x^{2} + 1} d x

\int \frac{(x^{2} + 1)}{x^{4} + 7 x^{2} + 1} d x =

\frac{(x^{2} - 5 x + 7) (x^{2} - x)}{(x + 1)} \geq 0

Join BYJU'S Learning Program