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ext{ extbackslash int extbackslash sqrt extbackslash left(}x^{ $2$ }+ $1$ ext{ extbackslash right)dx not using tangent}

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Find indefinite integrals using the substitution

Full solution

Q. ext{ extbackslash int extbackslash sqrt extbackslash left(}x^{

2

}+

1

ext{ extbackslash right)dx not using tangent}

Set $u = x^2 + 1$ : Let's set $u = x^2 + 1$ , then $du = 2x \, dx$ .
Express $dx$ in terms of $du$ : Now, we need to express $dx$ in terms of $du$ , so $dx = \frac{du}{2x}$ .
Substitute $u$ and $dx$ : Substitute $u$ and $dx$ into the integral: $\int\sqrt{x^2 + 1}\, dx = \int\sqrt{u} \cdot \left(\frac{du}{2x}\right)$ .
Express $x$ in terms of $u$ : We can't integrate with $x$ in the expression, so we need to express $x$ in terms of $u$ . Since $u = x^2 + 1$ , $x = \sqrt{u - 1}$ .
Substitute $x$ into integral: Now substitute $x$ into the integral: $\int \sqrt{u} \cdot \left(\frac{du}{2\sqrt{u - 1}}\right)$ .
Simplify the integral: Simplify the integral: $\int \frac{\sqrt{u}}{2\sqrt{u - 1}} \, du$ .
Try a substitution: This integral looks tough, let's try a substitution. Let's set $v = \sqrt{u}$ , then $v^2 = u$ and $2v \, dv = du$ .
Substitute $v$ and $du$ : Substitute $v$ and $du$ into the integral: $\int \frac{v}{2\sqrt{v^2 - 1}} \cdot 2v \, dv$ .
Simplify the integral: Simplify the integral: $\int \frac{v^2}{\sqrt{v^2 - 1}} \, dv$ .
Use the formula: This is a standard integral, we can use the formula $\int (\sec \theta)^2 \, d\theta = \tan \theta + C$ , where $\sec \theta = \frac{v}{\sqrt{v^2 - 1}}$ .
Correct the mistake: But we made a mistake, $\sec\theta$ should be $\sqrt{v^2 - 1}/v$ , not $v/\sqrt{v^2 - 1}$ . We need to correct this.

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