Let I = ∫x/√(x)+1dx Put √(x)=t⇒1/2√(x)dx=dt ⇒ dx = 2√(x)dt ∴ I = 2∫(x√(x)/t+1 )dt=2∫t^2.t/t + 1dt = 2∫t^3/t + 1dt = 2 ∫t^3+1 1/t + 1dt = 2 ∫(t+1)(t^2 ...

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Byju's Answer

Standard XII

Mathematics

Integration of Irrational Algebraic Fractions - 2

∫x/√x+1 d x

Question

$\int \frac{x}{\sqrt{x} + 1} d x$

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Solution

Let I = $\int \frac{x}{\sqrt{x} + 1} d x P u t \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow d x = 2 \sqrt{x} d t ∴ I = 2 \int (\frac{x \sqrt{x}}{t + 1}) d t = 2 \int \frac{t^{2} . t}{t + 1} d t = 2 \int \frac{t^{3}}{t + 1} d t = 2 \int \frac{t^{3} + 1 - 1}{t + 1} d t = 2 \int \frac{(t + 1) (t^{2} - t + 1)}{t + 1} d t - 2 \int \frac{1}{t + 1} d t = 2 \int (t^{2} - t + 1) d t - 2 \int \frac{1}{t + 1} d t = 2 [\frac{t^{3}}{3} - \frac{t^{2}}{2} + t - l o g | (t + 1) |] + C = 2 [\frac{x \sqrt{x}}{3} - \frac{x}{2} + \sqrt{x} - l o g | (\sqrt{x} + 1) |] + C$

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∫x√x+1dx

Let I = ∫x√x+1dxPut√x=t⇒12√xdx=dt⇒dx=2√xdt∴ I=2∫(x√xt+1)dt=2∫t2.tt+1dt=2∫t3t+1dt=2∫t3+1−1t+1dt=2∫(t+1)(t2−t+1)t+1dt−2∫1t+1dt=2∫(t2−t+1)dt−2∫1t+1dt=2[t33−t22+t−log|(t+1)|]+C=2[x√x3−x2+√x−log|(√x+1)|]+C

$\int \frac{x}{\sqrt{x} + 1} d x$