Question

If $I=\int \frac{x^3}{\sqrt{x^2+2x+5}}dx=\frac{1}{6}f\left(x\right)-\frac{A}{120}g\left(x\right)+C$ where $f\left(x\right)=(2x^2-5x-36)\sqrt{x^2+2x+5},g\left(x\right)=\log (x+1+\sqrt{x^2+2x+5})$. Then, A is equal to

Answer 1

$I=\int \frac{x^3}{\sqrt{x^2+2x+5}}dx=\int \frac{x^3}{\sqrt{{(x+1)}^2+4}}dx$
Substitute $t=x+1\Rightarrow dt=dx$
$I=\int \frac{{(t-1)}^3}{\sqrt{t^2+4}}dt$
Again substitute $t=2\tan u\Rightarrow dt=2{\sec }^{2}udu$
$I=\int {(2\tan u-1)}^3\sec udu=\int (-\sec u+8{\tan }^{3}u\sec u-12{\tan }^{2}u\sec u+6\tan u\sec u)du=8\int {\tan }^{3}u\sec udu-12\int {\tan }^{2}u\sec udu+6\int \tan u\sec udu-\int \sec udu=I_1+I_2+I_3+I_4$
Where
$I_1=8\int {\tan }^{3}u\sec udu=8\int \tan u\sec u({\sec }^{2}u-1)dup=\sec u\Rightarrow dp=\tan u\sec udu{I}_1=8\int (p^2-1)dp=-8p+\frac{8p^3}{3}=-8\sec u+\frac{8{\sec }^{3}u}{3}{I}_2=-12\int {\tan }^{2}u\sec udu=-12\int ({\sec }^{3}u-\sec u)du=-12\int {\sec }^{3}udu+12\int \sec udu=-6\tan u\sec u+6\int \sec udu=-6\tan u\sec u+6\log (\tan u+\sec u)I_3=6\int \tan u\sec uduw=\sec u\Rightarrow dw=\tan u\sec u{I}_3=\int dw=w=\sec u{I}_4=-\int \sec udu=-\log (\tan u+\sec u)$
Therefore,
$I=\frac{8}{3}\sec {\left({\tan }^{-1}\frac{t}{2}\right)}^3-6\tan \left({\tan }^{-1}\frac{t}{2}\right)\sec \left({\tan }^{-1}\frac{t}{2}\right)-2\sec \left({\tan }^{-1}\frac{t}{2}\right)+5\log (\tan \left({\tan }^{-1}\frac{t}{2}\right)+\sec \left({\tan }^{-1}\frac{t}{2}\right))=\frac{1}{6}(\sqrt{x^2+2x+5}(2x^2-5x-5)+30\log \left(\frac{1}{2}(x+\sqrt{{(x+1)}^2+4}+1)\right))$
Hence, $A=1200$