Иант ํo 15
$\int \frac{3 x + 5}{\sqrt{x^{2} + 4 x + 5}} d x$

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Answer

Re-substitute x: $x \sqrt{x^{2} + 4 x + 5} + \frac{9}{2} \ln (x^{2} + 4 x + 5) + C$

Steps

Step 1 ：Let $u = x^{2} + 4 x + 5$ , then $\frac{d u}{d x} = 2 x + 4$

Step 2 ：Separate the integral into two parts: $\int \frac{3 x + 5}{\sqrt{x^{2} + 4 x + 5}} d x = \int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 5}} d x + \int \frac{x + 1}{\sqrt{x^{2} + 4 x + 5}} d x$

Step 3 ：Now substitute u: $\int \frac{2 x + 4}{\sqrt{x^{2} + 4 x + 5}} d x = \frac{1}{2} \int \frac{d u}{\sqrt{u}}$ and $\int \frac{x + 1}{\sqrt{x^{2} + 4 x + 5}} d x = \int \frac{\frac{d u}{2} - 1}{\sqrt{u}}$

Step 4 ：Solve the integrals: $\frac{1}{2} \int \frac{d u}{\sqrt{u}} = \sqrt{u}$ and $\int \frac{\frac{d u}{2} - 1}{\sqrt{u}} = \frac{1}{2} u \sqrt{u} - \frac{9}{2} \ln (u)$

Step 5 ：Re-substitute x: $x \sqrt{x^{2} + 4 x + 5} + \frac{9}{2} \ln (x^{2} + 4 x + 5) + C$

Иант ํo 15∫3x+5x2+4x+5dx

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Иант ํo 15
$\int \frac{3 x + 5}{\sqrt{x^{2} + 4 x + 5}} d x$