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I need to solve the integral $$\int^x \frac{1}{t^2}e^{t^2}dt. $$ The answer should be $$\sqrt{\pi}\left(\text{erfi}(x)\right)-\frac{e^{x^2}}{x} + C$$

What does $\text{erfi}(x)$ mean? Can anybody explain how to solve this integral step-by-step? Thanks!

I need to solve the 2nd order differential equation $$y'' -2xy'+2y=0 $$ using that $y_1=Cx$. The solution of $y_2$ should then be \begin{align}y_2 &=y_1\int^x\frac{1}{{y_1(t)}^2}\exp \left(-\int^t{-2s}\,ds\right) \,dt\\ &=Cx\int^x\frac{1}{(Ct)^2}\exp \left(\int^t{(2s)}\,ds\right) \,dt\\ &=\frac{x}{C}\int^x\frac{1}{t^2}exp \left(\int^t{(2s)}\,ds\right) \,dt\\ &= \frac{x}{C}\int^x\frac{1}{t^2}e^{t^2}\, dt \end{align}

I need to solve the integral $$\int^x \frac{1}{t^2}e^{t^2}dt. $$ The answer should be $$\sqrt{\pi}\left(\text{erfi}(x)\right)-\frac{e^{x^2}}{x} + C$$

What does $\text{erfi}(x)$ mean? Can anybody explain how to solve this integral step-by-step? Thanks!

I need to solve the 2nd order differential equation $$y'' -2xy'+2y=0 $$ using that $y_1=Cx$. The solution of $y_2$ should then be \begin{align}y_2 &=y_1\int^x\frac{1}{{y_1(t)}^2}\exp \left(-\int^t{-2s}\,ds\right) \,dt\\ &=Cx\int^x\frac{1}{(Ct)^2}\exp \left(\int^t{(2s)}\,ds\right) \,dt\\ &=\frac{x}{C}\int^x\frac{1}{t^2} \exp \left(\int^t{(2s)}\,ds\right) \,dt\\ &= \frac{x}{C}\int^x\frac{1}{t^2}e^{t^2}\, dt \end{align}

I need to solve the integral $$\int^x \frac{1}{t^2}e^{t^2}dt $$ The answer should be $$\sqrt{\pi}\left(erfi(x)\right)-\frac{e^{x^2}}{x} + C$$

What does $erfi(x)$ mean? Can anybody explain how to solve this integral step-by-step? Thanks!

I need to solve the 2nd order differential equation $$y'' -2xy'+2y=0 $$ using that $y_1=Cx$. The solution of $y_2$ should then be \begin{gather}y_2=y_1\int^x\frac{1}{{y_1(t)}^2}\exp \left(-\int^t{-2s}\,ds\right) \,dt=Cx\int^x\frac{1}{(Ct)^2}\exp \left(\int^t{(2s)}\,ds\right) \,dt=\\ \frac{x}{C}\int^x\frac{1}{t^2}exp \left(\int^t{(2s)}\,ds\right) \,dt = \frac{x}{C}\int^x\frac{1}{t^2}e^{t^2}\, dt \end{gather}

I need to solve the integral $$\int^x \frac{1}{t^2}e^{t^2}dt. $$ The answer should be $$\sqrt{\pi}\left(\text{erfi}(x)\right)-\frac{e^{x^2}}{x} + C$$

What does $\text{erfi}(x)$ mean? Can anybody explain how to solve this integral step-by-step? Thanks!

I need to solve the 2nd order differential equation $$y'' -2xy'+2y=0 $$ using that $y_1=Cx$. The solution of $y_2$ should then be \begin{align}y_2 &=y_1\int^x\frac{1}{{y_1(t)}^2}\exp \left(-\int^t{-2s}\,ds\right) \,dt\\ &=Cx\int^x\frac{1}{(Ct)^2}\exp \left(\int^t{(2s)}\,ds\right) \,dt\\ &=\frac{x}{C}\int^x\frac{1}{t^2}exp \left(\int^t{(2s)}\,ds\right) \,dt\\ &= \frac{x}{C}\int^x\frac{1}{t^2}e^{t^2}\, dt \end{align}

I need to solve the integral $$\int^x \frac{1}{t^2}e^{t^2}dt $$. The answer should be $$\sqrt{\pi}\left(erfi(x)\right)-\frac{e^{x^2}}{x} + C$$

What does erfi(x) mean? Can anybody explain how to solve this integral step-by-step? Thanks!

I need to solve the 2nd order differential equation $$y'' -2xy'+2y=0 $$using that $y_1=Cx$. The solution of $y_2$ should then be $$y_2=y_1\int^x\frac{1}{{y_1(t)}^2}exp \left(-\int^t{-2s}ds\right) dt=Cx\int^x\frac{1}{(Ct)^2}exp \left(\int^t{(2s)}ds\right) dt=$$ $$ \frac{x}{C}\int^x\frac{1}{t^2}exp \left(\int^t{(2s)}ds\right) dt = \frac{x}{C}\int^x\frac{1}{t^2}e^{t^2} dt $$

I need to solve the integral $$\int^x \frac{1}{t^2}e^{t^2}dt $$ The answer should be $$\sqrt{\pi}\left(erfi(x)\right)-\frac{e^{x^2}}{x} + C$$

What does $erfi(x)$ mean? Can anybody explain how to solve this integral step-by-step? Thanks!

I need to solve the 2nd order differential equation $$y'' -2xy'+2y=0 $$  using that $y_1=Cx$. The solution of $y_2$ should then be \begin{gather}y_2=y_1\int^x\frac{1}{{y_1(t)}^2}\exp \left(-\int^t{-2s}\,ds\right) \,dt=Cx\int^x\frac{1}{(Ct)^2}\exp \left(\int^t{(2s)}\,ds\right) \,dt=\\ \frac{x}{C}\int^x\frac{1}{t^2}exp \left(\int^t{(2s)}\,ds\right) \,dt = \frac{x}{C}\int^x\frac{1}{t^2}e^{t^2}\, dt \end{gather}