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Question Number 167554 by cortano1 last updated on 19/Mar/22

$${\int }_0^{\mathrm{\infty }}\frac{{3\mathrm{x}}^2}{\sqrt{{({5\mathrm{x}}^2+1)}^3}}\mathrm{dx}$$$$$$

Answered by MJS_new last updated on 19/Mar/22

$$\int \frac{3x^2}{{(5x^2+1)}^{3/2}}dx=$$$$[t=\sqrt{5}x+\sqrt{5x^2+1}\to dx=\frac{\sqrt{5x^2+1}}{\sqrt{5}t}]$$$$=\frac{3\sqrt{5}}{25}\int \frac{{(t^2-1)}^2}{t{(t^2+1)}^2}dt=$$$$=\frac{3\sqrt{5}}{25}\int \frac{dt}{t}-\frac{12\sqrt{5}}{25}\int \frac{t}{{(t^2+1)}^2}dt=$$$$=\frac{3\sqrt{5}}{25}\ln t+\frac{6\sqrt{5}}{25(t^2+1)}=$$$$=\frac{3\sqrt{5}}{25}\ln (\sqrt{5}x+\sqrt{5x^2+1})-\frac{3x}{5\sqrt{5x^2+1}}+C$$

Commented by cortano1 last updated on 19/Mar/22

$$\mathrm{I}\mathrm{got}\mathrm{diverges}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{correct}?$$

Commented by MJS_new last updated on 19/Mar/22

$$\mathrm{yes}$$

Commented by cortano1 last updated on 19/Mar/22

$$\mathrm{thanks}\mathrm{sir}$$

Answered by greogoury55 last updated on 19/Mar/22

$$setx\sqrt{5}=\tan y$$$$I={\int }_0^{\frac{\pi }{2}}\frac{3\left(\frac{\tan {}^{2}y}{5}\right)}{{\left(\sqrt{\sec {}^{2}y}\right)}^3}\left(\frac{\sec {}^{2}y}{\sqrt{5}}\right)dy$$$$I=\frac{3}{5\sqrt{5}}{\int }_0^{\frac{\pi }{2}}\frac{\tan {}^{2}y}{\sec y}dy$$$$I=\frac{3}{5\sqrt{5}}{\int }_0^{\frac{\pi }{2}}\frac{1-\cos {}^{2}y}{\cos y}dy$$$$I=\frac{3}{5\sqrt{5}}{[\ln \mid \sec y+\tan y\mid -\sin y]}_0^{\frac{\pi }{2}}$$$$Diverges$$

Answered by Mathspace last updated on 20/Mar/22

$$\sqrt{5}x=tant\Rightarrow I=\frac{1}{\sqrt{5}}{\int }_0^{\frac{\pi }{2}}\frac{3{\left(\frac{tant}{\sqrt{5}}\right)}^2}{{(1+{tan}^2)}^{\frac{3}{2}}}(1+{tan}^{2}t)dt$$$$=\frac{3}{5\sqrt{5}}{\int }_0^{\frac{\pi }{2}}\frac{{tan}^{2}t}{\sqrt{1+{tan}^2}}dt$$$$=\frac{3}{5\sqrt{5}}{\int }_0^{\frac{\pi }{2}}\frac{1+{tan}^{2}t-1}{\sqrt{1+{tan}^{2}t}}dt$$$$=\frac{3}{5\sqrt{5}}{\int }_0^{\frac{\pi }{2}}\sqrt{1+{tan}^{2}t}dt-\frac{3}{5\sqrt{5}}{\int }_0^{\frac{\pi }{2}}\frac{dt}{\sqrt{1+{tan}^{2}t}}$$$$wehave{\int }_0^{\frac{\pi }{2}}\sqrt{1+{tan}^{2}t}dt$$$$={\int }_0^{\frac{\pi }{2}}\frac{dt}{cost}{=}_{tan\left(\frac{t}{2}\right)=y}{\int }_0^1\frac{2dy}{(1+y^2)\frac{1-y^2}{1+y^2}}$$$$={\int }_0^1\frac{2dy}{1-y^2}={\int }_0^1(\frac{1}{1-y}+\frac{1}{1+y})dy$$$$={[ln\mid \frac{1+y}{1-y}\mid ]}_0^1=\mathrm{\infty }$$$${\int }_0^{\frac{\pi }{2}}\frac{dt}{\sqrt{1+{tan}^2}}={\int }_0^{\frac{\pi }{2}}costdt$$$$={\left[sint\right]}_0^{\frac{\pi }{2}}=1sothisintegraleis$$$$divdrgente...!$$

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